Show that $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$
The cases $n=1$ and $n=2$ can be verified manually. We assume that $n\geq 3$. For an integer $k>2$, we have $$1+\frac{1}{k^3}=\left(1+\frac1k\right)\left(1-\frac1k+\frac1{k^2}\right)=\left(1+\frac1k\right)\left(\frac{k-1}{k}\right)^2\left(1+\frac{1}{k-1}+\frac{1}{(k-1)^2}\right).$$ We note that $$1+\frac{1}{k-1}+\frac1{(k-1)^2}=\frac{1-\frac{1}{(k-1)^3}}{1-\frac{1}{k-1}}<\frac{1}{1-\frac{1}{k-1}}=\frac{k-1}{k-2}$$ for $k>2$. That is, $$\prod_{k=1}^n\left(1+\frac1{k^3}\right)\leq \left(1+\frac1{1^3}\right)\left(1+\frac1{2^3}\right)\prod_{k=3}^n\left(1+\frac1k\right)\left(\frac{k-1}{k}\right)^2\left(\frac{k-1}{k-2}\right).$$ The RHS can be telescoped nicely: $$\prod_{k=3}^n\left(1+\frac1k\right)=\frac{n+1}{3},$$ $$\prod_{k=3}^n\left(\frac{k-1}{k}\right)=\frac{2}{n},$$ and $$\prod_{k=3}^n\left(\frac{k-1}{k-2}\right)=n-1.$$ Thus, $$\prod_{k=1}^n\left(1+\frac1{k^3}\right)\leq 2\left(\frac98\right)\left(\frac{n+1}{3}\right)\left(\frac{2}{n}\right)^2(n-1)=3\left(\frac{n^2-1}{n^2}\right)<3.$$
In fact, for a fixed positive integer $m\geq 3$ and for every $n\geq m$, we have $$\prod_{k=1}^n\left(1+\frac1{k^3}\right)\leq t_m\left(\frac{n^2-1}{n^2}\right) <t_m,$$ where $$t_m=\frac{m^2}{m^2-1}\ \prod_{k=1}^m\left(1+\frac{1}{k^3}\right).$$ If we pick $m=5$, we get $m=\frac{637}{256}<\frac{640}{256}=\frac52$. So, we can prove a stronger inequality $$\prod_{k=1}^n\left(1+\frac1{k^3}\right)<\frac52.$$
By writing $1+\frac{1}{k^3}$ as $$\frac{(k+1)(k-u)(k-v)}{k^3}=\frac{k+1}{k}\left(\frac{\Gamma(k+1-u)}{\Gamma(k-u)}\right)\left(\frac{\Gamma(k+1-v)}{\Gamma(k-v)}\right)\left(\frac{\Gamma(k)}{\Gamma(k+1)}\right)^2,$$ where $u=\frac{1+\sqrt{3}i}2$ and $v=\bar{u}=1-u$, we have $$\prod_{k=1}^n\left(1+\frac1{k^3}\right)=\frac{(n+1)\Gamma(n+1-u)\Gamma(n+1-v)}{\Gamma(1-u)\Gamma(1-v)\big(\Gamma(n+1)\big)^2}.$$ From the reflection formula, and from the relation $u+v=1$, we have $$\Gamma(1-u)\Gamma(1-v)=\Gamma(1-u)\Gamma(u)=\left(\frac{\pi}{\sin(\pi u)}\right).$$ Since \begin{align}\sin(\pi u)&=\sin\left(\frac{\pi}{2}+\frac{\sqrt{3}}{2}\pi i\right)\\&=\sin\left(\frac{\pi}{2}\right)\cosh\left(\frac{\sqrt{3}}{2}\pi\right)+i\cos\left(\frac{\pi}{2}\right)\sinh\left(\frac{\sqrt{3}}{2}\pi\right) \\&=\cosh\left(\frac{\sqrt{3}}{2}\pi\right),\end{align} we conclude that $$\Gamma(1-u)\Gamma(1-v)=\frac{\pi}{\cosh\left(\frac{\sqrt{3}}{2}\pi\right)}.$$ This leads to \begin{align}\prod_{k=1}^n\left(1+\frac1{k^3}\right)&=\frac{\cosh\left(\frac{\sqrt{3}}{2}\pi\right)}{\pi}\left(\frac{(n+1)\Gamma(n+1-u)\Gamma(n+1-v)}{\big(\Gamma(n+1)\big)^2}\right) \\&=\frac{\cosh\left(\frac{\sqrt{3}}{2}\pi\right)}{\pi}\left(\frac{\Gamma(n+1-u)\ (n+1)^u}{\Gamma(n+1)}\right)\left(\frac{\Gamma(n+1-v)\ (n+1)^v}{\Gamma(n+1)}\right).\end{align} Since $$\lim_{n\to\infty}\frac{\Gamma(n+z)}{\Gamma(n)\ n^z}=1$$ for all $z\in\mathbb{C}$, we get $$\prod_{k=1}^\infty\left(1+\frac1{k^3}\right)=\frac{\cosh\left(\frac{\sqrt{3}}{2}\pi\right)}{\pi}\approx 2.42818979.$$
Denote $p(n):=(1+\frac{1}{1^3})(1+\frac{1}{2^3})(1+\frac{1}{3^3})...(1+\frac{1}{n^3})$.
Claim: $$p(n)\leq3-\frac2{n^2},\,\forall n\geq2.$$
For $n=2$, we have $\frac94\leq3-\frac12$.
Then suppose $p(n)\leq3-\frac2{n^2}$ for some $n$. We see $$\eqalign{ p(n+1)&=p(n)(1+\frac1{(n+1)^3})\cr &\leq3-\frac{2}{n^2}+\frac3{(n+1)^3}-\frac{2}{n^2(n+1)^3}\cr &=3+\frac{3n^2-2(n^3+3n^2+3n+1)-2}{n^2(n+1)^3}\cr &=3-\frac{2n^3+3n^2+6n+4}{n^2(n+1)^3}\cr &=3-\frac{2n^3+2n^2+(n^2+6n+4)}{n^2(n+1)^3}\cr &\leq3-\frac{2n^2(n+1)}{n^2(n+1)^3}\cr &\leq3-\frac2{(n+1)^2}}.$$
As pointed out by @saulspatz, one can prove that $p(n)\leq3-\frac1n,\forall n\geq1$ by the same method.
Hope this helps.
We have that
$$\prod_{k=1}^\infty \left(1+\frac{1}{k^3}\right)<3\iff \sum_{k=1}^\infty \log\left(1+\frac{1}{k^3}\right)<\log 3$$
and since $\forall x>0\, \log(1+x)<x$
$$\sum_{k=1}^\infty \log\left(1+\frac{1}{k^3}\right)=\log 2+\sum_{k=2}^\infty \log\left(1+\frac{1}{k^3}\right)<\log 2+\sum_{k=2}^\infty \frac{1}{k^3}<\log 3$$