Yes, this follows by the de la Vallée-Poussin necessary and sufficient condition for the uniform integrability. Indeed, suppose that \begin{equation} \gamma n^2\to a \end{equation} (as $n\to\infty$) for some real $a\ge0$. Your integral is $$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma},$$ where the $X_i$'s are independent standard normal random variables. Introducing $N:=n(n-1)/2$, $X:=(X_1,\dots,X_n)$, and $\|X\|:=\sqrt{\sum_1^n X_i^2}$, and then using the arithmetic-geometric-mean inequality, we have $$\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma} \le\Big(\frac1N\,\sum_{1\le i<j\le n}|X_i-X_j|^2\Big)^{N\gamma} \\ =O\Big(\frac{\|X\|^2}n\Big)^{N\gamma}=O\Big(1+\frac{\|X\|^2}n\Big)^C $$ for $C:=a/2+1$ and all large enough $n$.
Note also that $\|X\|^2$ has the gamma distribution with parameters $n/2$ and $2$ and hence $E\|X\|^{2C}=O(n^C)$. So, $$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}=O(1)$$ and, similarly, $$E\Big[\Big(\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}\Big)^2\Big]=O(1).$$ Also, obviously, $t^2/t\to\infty$ as $t\to\infty$. So, we have the uniform integrability.
So, we only need to establish the convergence of
\begin{equation*}
{2\gamma}\sum_{1\le i<j\le n}\ln|X_i-X_j|=2\gamma NU_n
\end{equation*}
in probability, where
\begin{equation*}
U_n:=\frac1N\,\sum_{1\le i<j\le n}h(X_i,X_j)
\end{equation*}
is a so-called U-statistic with kernel $h(X_i,X_j):=\ln|X_i-X_j|$, and still $N=\binom n2=n(n-1)/2$. It is easy to see (cf. e.g. page 20) that $Var\,U_n=O(1/n)=o(1)$, whereas
$$EU_n=m:=E\ln|X_1-X_2|.$$
So, $U_n\to m$ in probability, whence
\begin{equation*}
2\gamma NU_n\to am
\end{equation*}
in probability and thus, by the uniform integrability,
$$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}\to e^{am}=\exp\{a\,E\ln|X_1-X_2|\}$$
as $n\to\infty$.