Step 1. If
$$I_1(n)=\sum_{1\leq k\leq\sqrt{n}}\left(\Gamma\left(\frac{k}{n}\right)\right)^{-k}$$
Then $\lim\limits_{n\to\infty}I_1(n)=0$.
Proof. Indeed, since $\Gamma$ is decreasing on $(0,1]$ we have
$$
I_1(n)\leq\sum_{1\leq k\leq\sqrt{n}}\left(\Gamma\left(\frac{1}{\sqrt{n}}\right)\right)^{-k}\leq\sum_{k=1}^\infty\left(\Gamma\left(\frac{1}{\sqrt{n}}\right)\right)^{-k}=\frac{1}{\Gamma(1/\sqrt{n})-1}$$
and step 1. follows.
Step 2. If
$$I_2(n)=\sum_{\sqrt{n}<k\leq n/2}\left(\Gamma\left(\frac{k}{n}\right)\right)^{-k}$$
Then $\lim\limits_{n\to\infty}I_2(n)=0$.
Proof. Recall that $\Gamma$ attains its minimum $\approx0.8856$, on $[1,2]$, at some some point $x_0\approx1.4616$. In particular, $\Gamma(x)\geq2/3$ for $1\leq x\leq 2$. So, for $\sqrt{n}<k\leq n/2$ we have
$$
\frac{k}{n}\Gamma\left(\frac{k}{n}\right)=\Gamma\left(1+\frac{k}{n}\right)
\geq\frac{2}{3}
$$
Thus, for $\sqrt{n}<k\leq n/2$, we have $\Gamma(k/n)>4/3$. It follows that
$$
I_2(n)\leq \sum_{k>\sqrt{n}}\left(\frac{3}{4}\right)^k=4\left(\frac{3}{4}\right)^{\lceil\sqrt{n}\rceil}
$$
and step 2. follows.
Step 3. If
$$I_3(n)=\sum_{n/2<k\leq n}\left(\Gamma\left(\frac{k}{n}\right)\right)^{-k}$$
Then $\lim\limits_{n\to\infty}I_3(n)=\dfrac{e^\gamma}{e^\gamma-1}$.
where $\gamma$ is the Euler-Mascheroni constant.
Proof.
Note first that, with $p=n-k$,
$$
I_3(n)=\sum_{0\leq p<n/2}\left(\Gamma\left(1-\frac{p}{n}\right)\right)^{p-n}
=\sum_{p=0}^\infty a_p(n)
$$
with
$$a_p(n)=\left\{\matrix{\left(\Gamma\left(1-\frac{p}{n}\right)\right)^{p-n}&\hbox{if}& 0\leq p<n/2\cr0&\hbox{otherwise}}\right.$$
Now, since $\Gamma(1)=1$ and $\Gamma'(1)=-\gamma$ we have, for a fixed $p$ and large $n$:
$$(p-n)\ln\Gamma\left(1-\frac{p}{n}\right)=(p-n)\ln\left(1+\frac{\gamma p}{n}+\mathcal{O}\left(\frac{1}{n^2}\right)\right)=-\gamma p+\mathcal{O}\left(\frac{1}{n}\right)$$
Thus
$$
\forall\,p\geq 0,\quad \lim_{n\to\infty}a_p(n)=e^{-\gamma p}.\tag{1}
$$
Now, we will need the next lemma.
Lemma. For $t\in[1/2,1]$ we have $(\Gamma(t))^{t/(1-t)}\geq \Gamma(1/2)=\sqrt{\pi}.$
Taking, this lemma for granted, we conclude by taking $t=1-p/n$ when $0\leq p<n/2$, that
$$
\forall\,p\geq 0,n\geq 1,\quad a_p(n)\leq \left(\frac{1}{\sqrt{\pi}}\right)^p.
\tag{2}
$$
and clearly, $$\sum_{p=0}^\infty \left(\frac{1}{\sqrt{\pi}}\right)^p<+\infty\tag{3}$$ Combining $(1)$, $(2)$ and $(3)$ we conclude that
$$
\lim_{n\to\infty}I_3(n)=\lim_{n\to\infty}\sum_{p=0}^\infty a_p(n)
=\sum_{p=0}^\infty\lim_{n\to\infty}a_p(n)=
\sum_{p=0}^\infty e^{-\gamma p}=\frac{e^\gamma}{e^\gamma-1}.$$
The desired conclusion follows:
$$
\lim_{n\to\infty}\sum_{1\leq k\leq n}\left(\Gamma\left(\frac{k}{n}\right)\right)^{-k}=
\lim_{n\to\infty}(I_1(n)+I_2(n)+I_3(n))=\frac{e^\gamma}{e^\gamma-1}.
$$
Proof of the Lemma. Let $f(t)=\dfrac{t}{1-t}\ln\Gamma(t)$.
Then $f'(t)=\dfrac{g(t)}{(1-t)^2}$ with
$$g(t)=\ln\Gamma(t)+t(1-t)\psi(t);\quad\hbox{where $\psi(t)=\Gamma'(t)/\Gamma(t)$}$$
and $g'(t)=(1-t)h(t)$ with
$$h(t)=2\psi(t)+t\psi'(t)$$
and finally $h'(t)=3\psi'(t)+t\psi''(t)=\sum_{k=0}^\infty\frac{3k+t}{(k+t)^3}>0$.
So, $h$ is increasing, and $\lim_{t\to0^+}h(t)=-\infty$, $h(1)=\frac{\pi^2}{6}-2\gamma>0$. This proves that $h(t)<0$ for $0<t<x_0$ and $h(t)>0$ for $x_0<t<1$, for some $x_0$.
And $g$ is decreasing on $[0,x_0]$ and increasing on $[x_0,1]$. But
$\lim_{t\to0^+}g(t)=+\infty$, $g(1)=0$. This proves that $g$ has exactly one change of sign on $(0,1)$ from positive to negative. This proves that the minimum of $f$ on $[1/2,1]$ is $\min(f(1/2),f(1))=f(1/2)$, and the lemma is proved.
