In a problem related to the study of the Weil-Petersson volume of the moduli space of bordered Riemann surfaces of genus $g$ with $m$ geodesic boundaries, all of length $\ell > 0$, I've encountered the following sum: $$f(g,m,\ell) \equiv \sum_{j=0}^{3g+m-2} \ell^j \sum_{|\boldsymbol{\alpha}| = j} \prod_{i=1}^m \frac{\alpha_i!}{(2\alpha_i)!},$$ where the $\alpha_i$ are positive integers, $|\boldsymbol{\alpha}| = \sum_{i=1}^m \alpha_i$ and $3g+m>2$. Since $g$ will be fixed I will abbreviate $f(g,m,\ell) \rightarrow f(m,\ell)$. I am most interested in the behavior of $f(m,\ell)$ as $m \rightarrow \infty$, and perhaps (but not necessarily) in the regime $\ell \ll 1$ (but fixed). In particular it would be useful to have bounds on $f$. A very simple upper bound (valid for all $g,m,\ell$) is $$f(m,\ell) \leq \sum_{j=0}^\infty \ell^j \sum_{|\boldsymbol{\alpha}| = j} \prod_{i=1}^m \frac{1}{\alpha_i!} = \sum_{j=0}^\infty \frac{(\ell m)^j}{j!} = e^{\ell m}.$$ Now I am wondering whether better bounds exist, perhaps when $l \ll 1$ is a small positive number. Clearly for any fixed $m$ we have $f(m,\ell) = 1 + \mathcal{O}(\ell)$ as $\ell \rightarrow 0$. For instance, might there be a $\ell_c$ such that for all $\ell < \ell_c$, $f(m,\ell) = 1 + o(1)$ as $m \rightarrow \infty$? If not, can we find an explicit lower bound $f(m,\ell) \geq g(m,\ell)$ for all $\ell$ and $m > m_0(\ell)$ with $g$ a diverging function as $m \rightarrow \infty$?
My attempts to far: What I have tried so far to tackle this problem is to formulate a (crude) lower bound starting from $$f(m,\ell) \geq \sum_{j=0}^{3g+m-2} \ell^j \sum_{|\boldsymbol{\alpha}| = j} \prod_{i=1}^m \frac{1}{(2\alpha_i)!} = \sum_{j=0}^{3g+m-2} \ell^j \frac{1}{2^m (2j)!} \sum_{k=0}^m \binom{m}{k}(m-2k)^{2j},$$ where this last sum was calculated by @metamorphy. With this I've attempted to proceed by approximating the $j$-sum in a perhaps not so rigorous fashion, namely $$\sum_{j=0}^{3g+m-2} \frac{\left[ \sqrt{\ell}(m-2k) \right]^{2j}}{(2j)!} \approx \sum_{j=0}^{\infty} \frac{\left[ \sqrt{\ell}(m-2k) \right]^{2j}}{(2j)!} = \cosh(\sqrt{\ell}(m-2k)).$$ I expect this approximation to hold because in the left-hand sum the terms in the series certainly become small when $j \gtrsim (e/2) \sqrt{\ell} m$ (for any $k$, and assuming $(e/2) \sqrt{\ell} m$ is large), which, when $\ell$ is small enough compared to 1, is much smaller than the highest index $j = 3g+m-2$ we are summing until. So I expect $$f(m,\ell) \overset{\approx}{\geq} \frac{1}{2^m} \sum_{k=0}^m \binom{m}{k} \cosh\left[ \sqrt{\ell} (m-2k) \right] \geq \frac{e^{\sqrt{\ell} m}}{2^{m+1}} \sum_{k=0}^m \binom{m}{k} \exp\left( -2 \sqrt{\ell} k \right) = \frac{\cosh^m(\sqrt{\ell})}{2},$$ which indeed diverges as $m \rightarrow \infty$ for any $\ell > 0$.
Too long for a comment.
Starting from @metamorphy's elegant solution, considering $$T_j=\frac{1}{2^m}\sum_{k=0}^m\binom{m}{k}(m-2k)^{2j}$$ what we can see is that $$T_j=m \, P_{j-1}(m)$$ and the first of these polynomials are $$\left( \begin{array}{cc} 1 & 1 \\ 2 & 3 m-2 \\ 3 & 15 m^2-30 m+16 \\ 4 & 105 m^3-420 m^2+588 m-272 \\ 5 & 945 m^4-6300 m^3+16380 m^2-18960 m+7936 \\ 6 & 10395 m^5-103950 m^4+429660 m^3-893640 m^2+911328 m-353792 \end{array} \right)$$
The coefficient of the largest power is $(2j-1)!!$ and the constant term seems to be the tangent numbers (look at sequence $A000182$ in $OEIS$)