We have
\begin{equation}
f'(x)=\frac{8 \alpha x}{n}+\psi\left(\frac{n}{2}-x+1\right)-\psi\left(\frac{n}{2}+x+1\right),
\end{equation}
where $\psi:=(\ln\Gamma)'$.
By the Gauss formula, Theorem 1.6.1, page 26
\begin{equation}
\psi(x)=\int_0^\infty\Big(\frac{e^{-z}}{z}-\frac{e^{-xz}}{1-e^{-z}}\Big)dz,
\end{equation}
we have
\begin{equation}
f'(x)=\frac{8 \alpha x}{n}-2\int_0^\infty\frac{dz}{1-e^{-z}}\,e^{-(1+n/2)z}\sinh xz.
\end{equation}
Since $\sinh$ is strictly convex on $(0,\infty)$, $f'$ is strictly concave on $(0,n/2)$. Also, $f'(0)=0$ and $f'(\frac n2)\to-\infty$ as $n\to\infty$.
So, eventually (i.e., for all large enough $n$), either $f'<0$ on $(0,n/2)$ (which will be the case if $f''(0)\le0$) or $f'$ changes sign only once on $(0,n/2)$, from $+$ to $-$, at some $x_n\in(0,n/2)$.
So, either $f$ is decreasing on $(0,n/2)$ (which will be the case if $f''(0)\le0$) or increasing-decreasing, attaining its only maximum at some $x_n\in(0,n/2)$.

As $x\downarrow0$,
\begin{equation}
f'(x)=\Big(\frac{8\alpha}{n}-2\psi'(n/2+1)\Big)x+O(x^3),
\end{equation}
and
\begin{equation}
\frac{8\alpha}{n}-2\psi'(n/2+1)=\frac{4\alpha-2}{n}+\frac2{n^2}+O(n^{-3})
\end{equation}
as $n\to\infty$.

So, if $\alpha<1/2$, then $f''(0)<0$ eventually, and so, $f$ is decreasing on $(0,n/2)$, with the maximum at $0$.

If $\alpha=1/2$, then $f'(c\sqrt n)=\left(4 c-\frac{16 c^3}{3}\right)
\left(\frac{1}{n}\right)^{3/2}+O\left(\left(\frac{1}{n}\right)^{5/2}\right)$ for each real $c>0$ as $n\to\infty$, and so, eventually $f$ attains its only maximum at some $x_n\sim c\sqrt n$, where $c=\sqrt3/2$.

Finally, if $\alpha>1/2$ and $c\in(0,1/2)$, then $f'(cn)=g(c)+O(1/n)$ as $n\to\infty$, where $g(c):=8 \alpha c+\ln \frac{1-2 c}{2 c+1}$ for each real $c>0$. We have $g(c)=0$ iff $\alpha=a(c):=-\frac1{8c}\,\ln\frac{1-2 c}{2 c+1}$. Note that $a(c)$ increases from $1/2$ to $\infty$ as $c$ increases from $0$ to $1/2$. So, eventually $f$ attains its only maximum at some $x_n\sim cn$, where $c=a^{-1}(\alpha)$.