First, we note that
$$
\sum\limits_{k = 0}^n {\frac{{( - 1)^k }}{{k!^2 (n - k)!}}} = \frac{{L_n (1)}}{{\Gamma (n + 1)}} = \sqrt {\frac{{\rm e}}{\pi }} \frac{1}{{n^{1/4} \Gamma (n + 1)}}\left( {\sin \left( {2\sqrt n + \frac{\pi }{4}} \right) + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right)} \right),
$$
where $L_n$ is the $n$th Laguerre polynomial (see, e.g., here). By the reflection formula
$$
\frac{1}{{(x - n)!}} = ( - 1)^{n + 1} \frac{{\sin (\pi x)}}{\pi }\Gamma (n - x).
$$
We also have
$$
\frac{{\Gamma (n + a)}}{{\Gamma (n + b)}} = n^{a - b} \left( {1 + \mathcal{O}\!\left( {\frac{1}{n}} \right)} \right)
$$
as $n\to +\infty$ with any fixed complex $a$ and $b$. See, for instance, here. Thus, the series in question converges iff the series
$$
\sum\limits_{n = 1}^\infty {\frac{1}{{n^{x + 5/4} }}\left( {\sin \left( {2\sqrt n + \frac{\pi }{4}} \right) + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right)} \right)}
$$
converges. I note here that
$$
\sum\limits_{n = 1}^N {\sin \left( {2\sqrt n + \frac{\pi }{4}} \right)} = - \sqrt N \cos \left( {2\sqrt N + \frac{\pi }{4}} \right) + \mathcal{O}(1),
$$
i.e., the Dirichlet test is not applicable. We can certainly say that the series converges absolutely whenever $\operatorname{Re}(x)>-\frac{1}{4}$.
Addendum. Using summation by parts and the above asymptotics for the partial sum of the sine terms, it can be verified that
$$
\sum\limits_{n = 1}^N {\frac{1}{{n^{x + 5/4} }}\sin \left( {2\sqrt n + \frac{\pi }{4}} \right)} = - \frac{1}{{N^{x + 3/4} }}\cos \left( {2\sqrt N + \frac{\pi }{4}} \right) + \mathcal{O}\!\left( {\frac{1}{{N^{x + 5/4} }}} \right).
$$
Thus the original series converges absolutely precisely when $\operatorname{Re}(x)>-\frac{3}{4}$.
