Having some work done, here is a refined version of my initial question.
For integer $m>0$ and $0<q<m/2$, consider the sum
  $$ S(m,q) = \sum_{i=0}^{m-q} \binom{m}{i} \binom{m-i}{q}^2. $$
I want to understand the behavior (as $m$ grows) of the quantity
  $$ \sigma(m) = \max_{0\le q\le m} \binom{m}{q}^{-1} S(m,q). $$

One can get pretty good estimates simply observing that
\begin{align*}
 \sum_{q=0}^m \binom{m}{q}^{-1} S(m,q) 
    &= \sum_{q=0}^m \sum_{i=0}^{m-q} \frac{(m-i)!(m-q)!}{i!q!((m-i-q)!)^2} \\
    &= \sum_{i+j\le m} \frac{(m-i)!(m-j)!}{i!j!((m-i-j)!)^2}.
\end{align*}
The right-hand side turns out to be a well-known sequence (OEIS A001906), asymptotically equal to $C\phi^{2m}$, with $\phi=(1+\sqrt5)/2$ and $C=\phi^2/\sqrt 5$. As a result,
  $$ \frac{(C+o(1))}m \phi^{2m}\le \sigma(m) \le (C+o(1))\,\phi^{2m}. $$
So, ultimately, my question is: What is the largest exponent, say $\tau$, such that
  $$ \sigma(m) < \frac{K}{m^\tau} \phi^{2m} $$
(with $K=K(\tau)$)?