Possible to get a closed form expression, or an upper bound, for $ f(n)=\sum_{m=1}^\infty \bigg(\frac{m+n}{3}\bigg)^{m+n}\bigg(\frac{1}{m}\bigg)^m$?

Only a comment.

I am not sure if you will like this, because my commentary shows that it's unlikely that you will get a satisfying answer in relation to a closed form .

$$\sum\limits_{m=1}^\infty\left(\frac{m+n}{3}\right)^{m+n}\left(\frac{1}{m}\right)^m = \frac{1}{3^n}\sum\limits_{v=0}^\infty{\binom n v}n^{n-v}\left(\frac{d^{v+1}}{dx^{v+1}}\int\limits_0^\infty\frac{h(e^{\frac{t}{3}e^{x-t}})^n}{1-\ln h(e^{\frac{t}{3}e^{x-t}})}dt\right)\big{|}_{x=0}$$

$h(x)~$ is the infinite power tower . $\,$It's useful if $~n\in\mathbb{N}_0$ .

Conclusion: Lower bound and an upper bound make sense, especially an asymptotic formula.

I think the upper bound $~\displaystyle n!\left(\frac{3}{2}\left(\frac{e}{2}\right)^n-\left(\frac{e}{3}\right)^n\right)~$ for $~f(n)~$ could be a good start for looking for a better one.

We can derive bounds and approximations for $f(n)$ that suggest properties that such a closed form must have.

$$f(n)=\sum_{i=1}^\infty\frac{\left(\frac{n+i}{3}\right)^{n+i}}{i^i} $$

When $n$ is fixed, the summands, $a_i$, are increasing over $(0,\frac{n}{2})$ but decreasing over $(\frac{n}{2},\infty)$. This can be shown by considering their derivatives. By bounding the summand with continuous curves and accounting for the peak at $n/2$, we can deduce that

$$f(n)\leq\int_{1/2}^{n/2-1/2}a_{i+1/2}\,\mathrm{d}i+ a_{n/2}+\int_{n/2+1/2}^{\infty}a_{i-1/2}\,\mathrm{d}i$$

Then, by substitution with $i\mapsto i\pm1/2$, $$f(n)\leq\int_{1}^{\infty}a_{i}\,\mathrm{d}i+ a_{n/2}$$

Rearranging $a_i$ to $\left(\frac{n+i}{3}\right)^{n}\cdot e^{-i\ln\left(\frac{3}{\frac{n}{i}+1}\right)}$ allows us to bound it by incorporating the inequality $1-\frac1x\leq \ln x \leq x-1$. Hence,

$$\left(\frac{n+i}{3}\right)^{n}\cdot e^{\frac{i\left(n-2i\right)}{n+i}}\leq a_i\leq \left(\frac{n+i}{3}\right)^{n}\cdot e^{\frac{n}{3}-\frac{2}{3}i}$$

Combining this upper bound with the previous integral inequality,

$$\begin{aligned}f(n)&\leq\frac{e^{\frac{n}{3}}}{3^{n}}\int_{1 }^{\infty }\frac{\left(n+i\right)^{n}}{e^{\frac{2}{3}i}}\,\mathrm{d}i+ \left(\frac{n}2\right)^{n} \\ &\leq \frac{3e^{n}}{2^{n+1}}\Gamma\left(n+1,\frac{2\left(n+1\right)}{3}\right)+ \left(\frac{n}2\right)^{n}\end{aligned}$$

where $\Gamma(s,x)=\int_1^\infty t^{s-1}e^{-t}\,\mathrm{d}t$ is the incomplete gamma function. When $x\gg 0$, we have $\Gamma(s,x)\approx x^{s-1}e^{-x}$. Hence, for large $n$, $$f(n)\approx \frac{3}{2e^{\frac{2}{3}}}\left(\frac{e^{\frac{1}{3}}}{3}\left(n+1\right)\right)^{n}+\left(\frac{n}{2}\right)^{n}$$

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Update 2019-11-05. Improved bounds.

Expanding on @user90369's ingenious use of Stirling's approximation to bound $n^n$, we can use the existing upper and lower bounds for $n!$ to sandwich $n^n$. We can then choose the relevant bounds for the numerator and denominator of the summands of $f(n)$ to find the improved bounds

$$\frac{\sqrt{2\pi}e^{n}}{3^{n}e}\leq \frac{f(n)}{\sum_{i=1}^{\infty}\frac{\left(n+i\right)!\sqrt{\frac{i}{n+i}}}{3^{n}\cdot i!}} \leq \frac{e^{n+1}}{\sqrt{2\pi}3^{n}} \\ \\ \frac{\sqrt{2\pi}}{e\sqrt{n+1}} \leq \frac{f(n)}{\frac{e^{n}}{3^{n}}\sum_{i=1}^{\infty}\frac{\left(n+i\right)!}{3^{i}\cdot i!}} \leq \frac{e}{\sqrt{2\pi}} \\ \frac{\sqrt{2\pi}}{e\sqrt{n+1}} \leq \frac{f(n)}{e^{n}n!\left(\frac{3}{2}\cdot2^{-n}-3^{-n}\right)} \leq \frac{e}{\sqrt{2\pi}}$$

So - unless I'm mistaken, since I'm not great with asymptotic analysis - this means that $f(n)\in\mathcal{O}\left(\sqrt{n}\left(\frac{n}2\right)^n\right)$ and $f(n)\in\mathcal{\Omega}\left(\left(\frac{n}2\right)^n\right)$. This lends itself to the natural approximation $f(n)\approx k\sqrt{n}\left(\frac{n}{2}\right)^n$ or $f(n)\approx k\left(\frac{n}{2}\right)^n$, the best of which seems to be the good approximation

$$f(n)\approx 2.172\sqrt{n}\left(\frac{n}{2}\right)^n$$