What are better approximations to $\pi$ as algebraic though irrational number?

(expanding my comments)

Let's start with the fraction $\;\dfrac{355}{113}\,$ easy to remember with something like :
"doubling the odds to be near the pi" (whatever this may mean...).

It is easy to find starting with the continued fraction of $\pi$ and stopping just before the (relatively) large term $292$ : \begin{align} \pi&=[3; 7, 15, 1\color{#00ff00}{, 292, 1, 1, 1, 2, 1, 3, 1, 14,\cdots}]\\ \pi&\approx \frac{355}{113}\approx 3.141592\color{#808080}{035}\\ \end{align}

My next step will be to compute the continued fractions of the first powers of $\pi\,$ and stop the expansion before the first large term (as previously) to get :

$$\frac {355}{113},\; \sqrt{\frac{227}{23}},\; \sqrt[3]{31},\;\sqrt[4]{\frac{2143}{22}},\;\sqrt[5]{306},\cdots, \;\sqrt[11]{294204},\cdots$$

After the first power the most interesting term was the fourth : \begin{align} \pi^4&=[97; 2, 2, 3, 1\color{#00ff00}{, 16539, 1, 6, 7, 6,\cdots}]\\ \pi^4&\approx \frac{2143}{22}\\ \pi&\approx \sqrt[4]{\frac{2143}{22}}\approx 3.14159265\color{#808080}{258}\\ \end{align}

Mnemonic : think at "three ways to reverse two two" that I'll note $\pi\approx \sqrt{+ \negthickspace+/}$ :

1. Power way: two $\sqrt{}\;$ to reverse the double squaring $^2$ $^2$.
2. Incremental way : reverse two times two terms of the $2\times 2$ terms $\;\underbrace{12}\underbrace{34}$
3. Divide by $\,22$.

This solution is interesting because of the large (omitted) $16539$. Should we incorporate this term in the c.f. then the next numerator and denominator would have around $4$ additional digits (since $\log_{10}(16539)\approx 4.2\;$ and from the method to obtain the next fraction in the first link).
The precision will be better with this supplementary term (say $4.3$ digits more) but we needed $4+4$ more digits for this. Without this term we used $4+2=6$ digits for a result of $10$ digits (excellent), with this term we have $8+6=14$ digits for a result of $14$ digits (average for a c.f.).

Searching the largest terms at the beginning of a c.f. (excluding the first non-zero term) should thus be rather interesting! Unfortunately c.f. coefficients as large as $16539$ are rather uncommon.

This result was found by Ramanujan and is given too by Mathworld with many others. $$-$$ Some additional results :

A palindrome for the fractional part of $\pi$ : $\frac 1{\large{\sqrt[5]{17571}}}\approx 0.1415926\color{#808080}{48}$ (with two more terms this becomes $\sqrt[5]{\dfrac{296}{5201015}}\approx 0.141592653589\color{#808080}{63}$). Another one : $\;\dfrac 1{\sqrt[8]{6189766}} \approx 0.141592653\color{#808080}{64}$.

We may too search continued fractions $\dfrac{\log\pi}{\log n}\,$ to obtain : \begin{align} 7^{10/17}&\approx 3.141\color{#808080}{35}\\ 6^{23/36}&\approx 3.1416\color{#808080}{09}\\ 7^{58701/99785}&\approx 3.1415926535\color{#808080}{9651}\\ \end{align}

Other random solutions perhaps nearer to OP's question (with some usual c.f. for reference) : \begin{align} \frac{22}7 &\approx 3.14\color{#808080}{2857}\\ \frac{8.5^2}{23} &\approx 3.141\color{#808080}{30}\\ \sqrt[3]{31}&\approx 3.141\color{#808080}{38}\\ \sqrt{51}-4 &\approx 3.141\color{#808080}{428}\\ \sqrt{4508}-64 &\approx 3.141\color{#808080}{64}\\ 4-\sqrt{\frac {14}{19}} &\approx 3.141\color{#505050}{60}\color{#808080}{49}\\ 7-\left(\frac{55}{28}\right)^2 &\approx 3.1415\color{#808080}{816}\\ 1+\left(\frac{60}{41}\right)^2 &\approx 3.1415\color{#808080}{82}\\ \sqrt{14434}-117 &\approx 3.1415\color{#808080}{83}\\ 2+\sqrt[17]{9.5} &\approx 3.14159\color{#808080}{78}\\ 5-\sqrt[5]{22+\frac{1}6} &\approx 3.14159\color{#808080}{62}\\ \sqrt{\frac{1961}2}-19 &\approx 3.1415\color{#808080}{898}\\ 2+\sqrt[8]{\frac{75}{26}} &\approx 3.141592\color{#808080}{19}\\ \frac{355}{113} &\approx 3.141592\color{#808080}{92}\\ \sqrt[11]{294204} &\approx 3.1415926\color{#808080}{36}\\ \left(\sqrt{\frac {1731}{76}}-3\right)^2 &\approx 3.1415926\color{#808080}{65}\\ \sqrt{6}+\sqrt[3]{\frac {61}{184}}&\approx 3.1415926\color{#808080}{45}\\ \sqrt{35}-\sqrt[3]{\frac{6215}{291}} &\approx 3.14159265\color{#808080}{266}\\ \sqrt[4]{\frac{2143}{22}}&\approx 3.14159265\color{#808080}{258}\\ 5-\sqrt[11]{913+\frac 16} &\approx 3.141592653\color{#808080}{37}\\ \sqrt{5}+\sqrt[4]{\frac{2323}{3455}} &\approx 3.141592653\color{#808080}{436}\\ \sqrt{4508-\frac 1{153}}-64 &\approx 3.1415926535\color{#808080}{28}\\ \sqrt[4]{\frac{788453}{95}}-\sqrt{41} &\approx 3.1415926535\color{#808080}{918} \\ \sqrt[4]{\sqrt{\frac{1087906}{63}}-34}&\approx 3.14159265358\color{#808080}{876}\\ \frac{5419351}{1725033}&\approx 3.141592653589\color{#808080}{815}\\ \sqrt{7}+\sqrt[8]{\frac{94680}{25912921}} &\approx 3.141592653589793\color{#808080}{309}\\ \sqrt{\sqrt{\frac{10521363651}{311209}}-174} &\approx 3.141592653589793238\color{#808080}{01}\\ \frac{21053343141}{6701487259}&\approx 3.141592653589793238462\color{#808080}{38}\\ \sqrt{\sqrt{\frac{20448668456155}{3958899937}}-62} &\approx 3.14159265358979323846264338\color{#808080}{5}\\ \sqrt{12}-\sqrt[3]{\frac{626510899334}{18676834489131}} &\approx 3.1415926535897932384626433832\color{#505050}{80}\color{#808080}{4} \end{align}

We could too use the integer relation algorithms as in Will Jagy's answer or this one but this seems more cumbersome for this problem.

If you want to stay with degree two or three but no larger, find an implementation of PSLQ and feed it the quadruple (at incredible decimal accuracy) $$ \left(\pi^3, \; \pi^2, \; \pi, \; 1 \right) $$ so as to ask for integer relations, that is integers $a_3, a_2, a_1, a_0$ of not terribly large absolute value, so that $$ a_3 \pi^3 + a_2 \pi^2 + a_1 \pi + a_0 $$ is very close to zero. Then the relevant root of $a_3 x^3 + a_2 x^2 + a_1 x + a_0$ is a good approximation for $\pi.$

The others appear to be getting good results with degree four, you might try that, no more difficult once you have the code for the general cubic correct.

jagy@phobeusjunior:~$ gp
Reading GPRC: /etc/gprc ...Done.

                               GP/PARI CALCULATOR Version 2.5.5 (released)
                        i686 running linux (ix86/GMP-5.1.2 kernel) 32-bit version
                    compiled: Sep 30 2013, gcc-4.8.1 (Ubuntu/Linaro 4.8.1-10ubuntu4) 
                 (readline v6.3 enabled [was v6.2 in Configure], extended help enabled)

                                 Copyright (C) 2000-2013 The PARI Group

PARI/GP is free software, 
? Pi
%6 = 3.141592653589793238462643383
? q = algdep(Pi,4)
%7 = 5871*x^4 - 22872*x^3 - 7585*x^2 + 60199*x + 23027
? polroots(q)
%8 = [-1.311564323926921157096862611 + 0.E-28*I, -0.3879438664397374306161177256 + 0.E-28*I, 
2.453674351288873525029590438 + 0.E-28*I, 
3.141592653589793238462643859 + 0.E-28*I]~
? 

degrees five to ten

?  algdep(Pi,5)
%19 = 909*x^5 - 3060*x^4 + 1814*x^3 - 3389*x^2 - 723*x - 626
?  algdep(Pi,6)
%20 = 820*x^6 - 2340*x^5 - 565*x^4 + 67*x^3 - 1782*x^2 - 1008*x + 1460
?  algdep(Pi,7)
%21 = 306*x^7 - 1189*x^6 + 532*x^5 + 224*x^4 + 899*x^3 + 474*x^2 + 389*x + 485
?  algdep(Pi,8)
%22 = 27*x^8 + 46*x^7 - 256*x^6 - 564*x^5 + 43*x^4 + 672*x^3 - 104*x^2 - 201*x + 220
?  algdep(Pi,9)
%23 = 20*x^9 - 53*x^8 + 32*x^7 - 178*x^6 - 86*x^5 - 11*x^4 + 142*x^3 + 410*x^2 + 34*x + 21
?  algdep(Pi,10)
%24 = 2*x^10 - 5*x^9 - 17*x^8 + 47*x^7 - 64*x^6 + 146*x^5 - 58*x^4 + 79*x^3 + 110*x^2 + 23*x - 7
? 

degree three:

    ?  r = algdep(Pi,3)
    %26 = 91273*x^3 + 8437*x^2 - 960500*x + 104194
    ? polroots(r)
    %27 = [-3.342734408288101386537745201 + 0.E-28*I, 0.1087047799083921816885401406 + 0.E-28*I, 
3.141592653589793238462650438 + 0.E-28*I]~
    ? 
    ? 

degree two:

?  s = algdep(Pi,2)
%28 = 12610705*x^2 - 51111434*x + 36108636
? polroots(s)
%29 = [0.9114269040003652816200798826 + 0.E-28*I, 3.141592653589793238462659346 + 0.E-28*I]~

repeating degree ten, I like how the coefficients are small and begin with 2, I have not found any of these monic (beginning with $1$)

?  t = algdep(Pi,10)
%30 =   2*x^10 - 5*x^9 - 17*x^8 + 47*x^7 - 64*x^6 + 
       146*x^5 - 58*x^4 + 79*x^3 + 110*x^2 + 23*x - 7
? polroots(t)
%31 = [-3.416642530754670637725737702 + 0.E-28*I,
        0.1631777144832237629669559802 + 0.E-28*I, 
        2.659776825745310085407479343 + 0.E-28*I, 
        3.141592653589793238462643332 + 0.E-28*I,
       -0.4285725799568636122958113382 - 0.1971284716837764691749795140*I,
       -0.4285725799568636122958113382 + 0.1971284716837764691749795140*I,
        0.6277749736794889930752953905 - 1.073388946479318133923381580*I, 
        0.6277749736794889930752953905 + 1.073388946479318133923381580*I, 
       -0.2231547252544536053351545286 - 1.460683263806221846450712438*I, 
       -0.2231547252544536053351545286 + 1.460683263806221846450712438*I]~
? 

pretty graph:

How about,

$$ \sqrt[3] {31}=3.14138...$$

Where, $31$ is the length of a month.

If you want memorable, you could always use,

$$\pi \sim \sqrt{{{69} \over {7}}}=3.139...$$

Do I really need to explain this one?

You could also use,

$$\sqrt{{69 \cdot 1001} \over {7 \cdot 1000}}=3.14117...$$

Where, $1001$ refers to the book 1001 Arabian Nights