Computing $\lim_{n \to \infty} \left[\left(\prod_{i=1}^{n}i!\right)^{1\over n^{2}} (n^{x})\right] $ if exists for certain $x\in\mathbb R$
Asymptotic Expansion via Riemann Sum
Compute the log of the product as a Riemann Sum $$ \begin{align} \frac1{n^2}\sum_{k=1}^n k\log(n-k+1) &=\sum_{k=1}^n\frac{k}{n}\left(\log\left(1-\frac{k}{n}+\frac1n\right)+\log(n)\right)\frac1n\tag{1a}\\ &\sim\int_0^1x\log(1-x)\,\mathrm{d}x+\frac12\log(n)\tag{1b}\\ &=\int_0^1\log(1-x)\,\mathrm{d}\frac{x^2-1}2+\frac12\log(n)\tag{1c}\\ &=-\int_0^1\frac{x+1}2\,\mathrm{d}x+\frac12\log(n)\tag{1d}\\ &=\frac12\log(n)-\frac34\tag{1e} \end{align} $$ Thus, the product is asymptotically $$ \left(\prod_{k=1}^nk!\right)^{1/n^2}\sim e^{-3/4}n^{1/2}\tag2 $$ Therefore, for $x=-1/2$, the limit comes to $$ \bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}\left(\prod_{k=1}^nk!\right)^{1/n^2}n^{-1/2}=e^{-3/4}}\tag3 $$ For $x\lt-1/2$, the limit is $0$.
Asymptotic Expansion via Euler-Maclaurin Sum Formula
As is shown in this answer, we have asymptotically in $n$, $$ \sum_{k=1}^n k^{-z} =\zeta(z)+\frac{n^{1-z}}{1-z}+\frac12n^{-z}-\frac{z}{12}n^{-1-z}+O\left(\frac1{n^{3+z}}\right)\tag4 $$ applying $-\frac{\mathrm{d}}{\mathrm{d}z}$: $$ \begin{align} \sum_{k=1}^n\log(k)k^{-z} &=-\zeta'(z)+n^{1-z}\frac{(1-z)\log(n)-1}{(1-z)^2}+\frac12\log(n)n^{-z}\\ &-n^{-1-z}\frac{z\log(n)-1}{12}+O\!\left(\frac{\log(n)}{n^{3+z}}\right)\tag5 \end{align} $$ Setting $z=0$: $$ \sum_{k=1}^n\log(k)=\overbrace{\,\,-\zeta'(0)\ }^{\frac12\log(2\pi)}+n(\log(n)-1)+\frac12\log(n)+\frac1{12n}+O\!\left(\frac{\log(n)}{n^3}\right)\tag6 $$ Setting $z=-1$: $$ \begin{align} \sum_{k=1}^n\log(k)k &=\overbrace{-\zeta'(-1)}^{\log(A)-\frac1{12}}+n^2\frac{2\log(n)-1}4+\frac12n\log(n)+\frac{\log(n)+1}{12}\\ &+O\!\left(\frac{\log(n)}{n^2}\right)\tag7 \end{align} $$ where $A$ is the Glaisher–Kinkelin constant.
Thus, $$ \begin{align} \sum_{k=1}^n(n-k+1)\log(k) &=n^2\frac{2\log(n)-3}4+n\log\left(\frac{\sqrt{2\pi}}en\right)+\frac5{12}\log(n)\\ &+\log\left(\frac{\sqrt{2\pi}}{A}\right)+\frac1{12}+\frac1{12n}+O\!\left(\frac{\log(n)}{n^2}\right)\tag8 \end{align} $$ and therefore, $$ \prod_{k=1}^nk!=\frac{\sqrt{2\pi}}{A}e^{1/12}\,\color{#C00}{n^{n^2/2}e^{-3n^2/4}}\color{#090}{\left(\frac{\sqrt{2\pi}}en\right)^n}n^{5/12}\color{#00F}{e^{\frac1{12n}+O\left(\frac{\log(n)}{n^2}\right)}}\tag9 $$ Finally, $$ \bbox[5px,border:2px solid #C0A000]{\left(\prod_{k=1}^nk!\right)^{1/n^2}=\color{#C00}{n^{1/2}e^{-3/4}}+\color{#090}{O\!\left(\frac{\log(n)}{n^{1/2}}\right)}}\tag{10} $$
The Glaisher–Kinkelin Constant
Equation $(6)$ is essentially Stirling's Formula: $$ \prod_{k=1}^nk=\sqrt{2\pi}\,n^{n+1/2}e^{-n}\left(1+\frac1{12n}+O\!\left(\frac1{n^2}\right)\right)\tag{11} $$ where $\sqrt{2\pi}=e^{-\zeta'(0)}$. This shows that $\zeta'(0)=-\frac12\log(2\pi)$.
Equation $(7)$ says that $$ \prod_{k=1}^nk^k=A\,n^{n^2/2+n/2+1/12}e^{-n^2/4}\left(1+O\!\left(\frac{\log(n)}{n^2}\right)\right)\tag{12} $$ where $A=e^{\frac1{12}-\zeta'(-1)}$.
Just as $$ \sum_{k=1}^n\frac1k=\log(n)+\gamma+O\!\left(\frac1n\right)\tag{13} $$ is the defining limit for $\gamma$, the Euler-Mascheroni Constant, $(12)$ appears to be the defining limit for $A$, the Glaisher–Kinkelin constant.
Here is a more accurate than needed estimate from my answer at Formula for $1! \times 2! \times \cdots \times n!$?:
$\prod\limits_{k=0}^{n} n! \sim C^{1/2} (2\pi)^{3/8}n^{5/12}(2\pi)^{n/2}(n/e)^n \left(\dfrac{n}{e^{3/2}}\right)^{n^2/2} $ where $C =\lim\limits_{n \to \infty} \dfrac1{n^{1/12}}\prod\limits_{k=1}^n\left( \dfrac{k!}{\sqrt{2\pi k}(k/e)^k} \right) \approx 1.046335066770503 $.
Taking the $n^2$ root, $\left(\prod\limits_{k=0}^{n} n!\right)^{1/n^2} \to\left(\dfrac{n}{e^{3/2}}\right)^{1/2} $ since all the other terms go to $1$.