By the triangular inequality, $|e_{n,k}|\leqslant\frac12 a_n$ with $ a_n=\sum_{m=1}^nb_{n,m},\qquad b_{n,m}=\frac{(m!)^2(n-m)!}{m^3(n+m)!}, $ hence the desired uniform convergence holds as soon as $a_n\to0$.
To prove that $a_n\to0$, note that $ b_{n,m}=\frac{(m-1)!(m-1)!(n-m)!}{m(n+m)!}\leqslant\frac{(m-1)!(m-1)!(n-m)!}{(n+m)!}, $ and use twice the fact that $i!j!\leqslant (i+j)!$ for every nonnegative integers $i$ and $j$. This yields $ b_{n,m}\leqslant\frac{(n+m-2)!}{(n+m)!}=\frac1{(n+m-1)(n+m)}=\frac1{n+m-1}-\frac1{n+m}. $ Thus $a_n$ is bounded by a telescoping sum, namely $ a_n\leqslant\sum_{m=1}^n\left(\frac1{n+m-1}-\frac1{n+m}\right)=\frac1{n}-\frac1{2n}=\frac1{2n}, $ This shows that $|e_{n,k}|\leqslant\frac1{4n}$ for every $n\geqslant1$ and uniformly over $k\leqslant n$, hence the proof is complete.
Edit One can refine the estimates above, taking into account the variations of the sequence $(b_{n,m})_{1\leqslant m\leqslant n}$ for some fixed $n$, which is nonincreasing on $1\leqslant m\leqslant m_n$ for some $m_n\approx n/\sqrt2$ and nondecreasing on $m_n\leqslant m\leqslant n$. Thus, $|e_{n,k}|\leqslant\frac12 b_{n,1}+\frac12 b_{n,n}$. Hence, $n^2e_{n,k}\to\frac12 $ uniformly over $k$.