A closed form of $\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\Gamma^2\left(\frac{k}{2}\right)$
You can use the identity given by the Euler Beta function $$\int_{0}^{1}x^{a-1} (1-x)^{b-1} \,dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ to state: $$S=\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k!}\Gamma(k/2)^2=\sum_{k=1}^{+\infty}\frac{(-1)^{k-1}}{k}\int_{0}^{1}\left(x(1-x)\right)^{k/2-1}\,dx $$ and by switching the series and the integral: $$ S = \int_{0}^{1}\frac{\log(1+\sqrt{x(1-x)})}{x(1-x)}dx = 2\int_{0}^{1/2}\frac{\log(1+\sqrt{x(1-x)})}{x(1-x)}dx,$$ $$ S = 2\int_{0}^{1/2}\frac{\log(1+\sqrt{1/4-x^2})}{1/4-x^2}dx = 4\int_{0}^{1}\frac{\log(1+\frac{1}{2}\sqrt{1-x^2})}{1-x^2}dx,$$ $$ S = 4\int_{0}^{\pi/2}\frac{\log(1+\frac{1}{2}\sin\theta)}{\sin\theta}d\theta.$$ Now Mathematica gives me $\frac{5\pi^2}{18}$ as an explicit value for the last integral, but probably we are on the wrong path, and we only need to exploit the identity $$\sum_{k=1}^{+\infty}\frac{1}{k^2\binom{2k}{k}}=\frac{\pi^2}{18}$$ that follows from the Euler acceleration technique applied to the $\zeta(2)$-series. The other "piece" (the $U$-piece in the Marty Cohen's answer) is simply given by the Taylor series of $\arcsin(z)^2$. More details to come.
As a matter of fact, both approaches lead to an answer. The (Taylor) series approach, as Bhenni Benghorbal shows below, leads to the identity: $$\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\Gamma^2\left(\frac{k}{2}\right)x^k= 2 \arcsin \left( x/2 \right) \left(\pi - \arcsin \left( x/2\right) \right),\tag{1}$$ while the integral approach, as Achille Hui pointed out in the comments, leads to: $$\begin{eqnarray*}\int_{0}^{\pi/2}\frac{\log(1+\frac{1}{2}\sin\theta)}{\sin\theta}\,d\theta&=&\int_{0}^{1}\log\left(1+\frac{t}{1+t^2}\right)\frac{dt}{t}\\&=&\int_{0}^{1}\frac{\log(1-t^3)-\log(1-t)-\log(1+t^2)}{t}\,dt\\&=&\int_{0}^{1}\frac{-\frac{2}{3}\log(1-t)-\frac{1}{2}\log(1+t)}{t}\,dt\\&=&\frac{1}{6}\sum_{k=1}^{+\infty}\frac{4+3(-1)^k}{k^2}=\frac{1}{6}\left(4-\frac{3}{2}\right)\zeta(2)=\frac{5\pi^2}{72}.\end{eqnarray*}\tag{2}$$
Thanks to both since now this answer may become a reference both for integral-log-ish problems like $(2)$ and for $\Gamma^2$-series like $(1)$.
Update 14-06-2016. I just discovered that this problem can also be solved by computing $$ \int_{-1}^{1} x^n\, P_n(x)\,dx, $$ where $P_n$ is a Legendre polynomial, through Bonnet's recursion formula or Rodrigues' formula. Really interesting.
$($This is more of a comment than an answer, but $...)$
Consider the even and odd $k$ in separate sums:
Note: You probably do not want $\Gamma(0)$ in the sum, so I'll start at $k=1$.
$$\begin{align} \\S &=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\left[\Gamma\left(\frac{k}{2}\right)\right]^2 \\\\ &=\sum_{k=1}^\infty\frac{(-1)^{2k+1}}{(2k)!}\left[\Gamma\left(\frac{2k}{2}\right)\right]^2 ~+~\sum_{k=0}^\infty\frac{(-1)^{2k+1+1}}{(2k+1)!}\left[\Gamma\left(\frac{2k+1}{2}\right)\right]^2 \\\\ &=\sum_{k=1}^\infty\frac{-1}{(2k)!}\left[\Gamma\left(k\right)\right]^2 ~+~\sum_{k=0}^\infty\frac{1}{(2k+1)!}\left[\Gamma\left(k+\frac12\right)\right]^2 \\\\ &=-\sum_{k=1}^\infty\frac{((k-1)!)^2}{(2k)!} ~+~\sum_{k=0}^\infty\frac{1}{(2k+1)!}\left[\frac{\sqrt{\pi}(2k)!}{4^kk!}\right]^2 \\\\ &=-\sum_{k=1}^\infty\frac{((k-1)!)^2}{(2k)!} ~+~\pi\sum_{k=0}^\infty\frac{(2k)!}{(2k+1)16^k(k!)^2} \\\\ &=-T~+~\pi~U, \\\\ \end{align}$$
where $~T=\displaystyle\sum_{k=1}^\infty\frac{((k-1)!)^2}{(2k)!},~$ and $~U=\displaystyle\sum_{k=0}^\infty\frac{(2k)!}{(2k+1)16^k(k!)^2}.$
You can write these sums in terms of the central binomial coefficients $\displaystyle\binom{2k}{k}=\frac{(2k)!}{(k!)^2},$
and people who know more about these than I can possibly sum these.
I'll leave it at this.
Here is a closed form
$$ \mathcal{I}=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\Gamma^2\left(\frac{k}{2}\right)x^k= 2 \arcsin \left( x/2 \right) \left(\pi - \arcsin \left( x/2\right) \right) .$$
Now just plug in $x=1$ and the result follows.