Binomial coefficients have a well known asymptotics (https://en.wikipedia.org/wiki/Binomial_coefficient#Bounds_and_asymptotic_formulas) given by $$\binom nk\sim\binom{n}{\frac{n}{2}} e^{-d^2/(2n)} \sim \frac{2^n}{\sqrt{\frac{1}{2}n \pi }} e^{-d^2/(2n)}.$$
Is there corresponding asymptotics for multinomials (I am more interested in what happens analogously in the denominator)?
Suppose that $k$ is a fixed natural number, $n\to\infty$, and
\begin{equation*}
a_i=\frac nk+o(n^{2/3})
\end{equation*}
for each $i$; here in what follows, $i\in\{1,\dots,k\}$.
Let
\begin{equation*}
h_i:=\frac kn\,a_i-1=o(n^{-1/3}), \tag{1}
\end{equation*}
so that $h_i\to0$ and
\begin{equation}
a_i=\frac nk\,(1+h_i). \tag{2}
\end{equation}
By Stirling's formula,
\begin{equation*}
n!\sim(2\pi)^{1/2} n^{1/2}\Big(\frac ne\Big)^n
\end{equation*}
and, by (2),
\begin{equation*}
\begin{aligned}
a_i!&\sim(2\pi)^{1/2} \Big(\frac nk\Big)^{1/2}(1+h_i)^{1/2}\Big(\frac n{ke}(1+h_i)\Big)^{a_i} \\
&\sim(2\pi)^{1/2} \Big(\frac nk\Big)^{1/2}\Big(\frac n{ke}(1+h_i)\Big)^{a_i}.
\end{aligned}
\end{equation*}
Therefore and because $\sum_{i=1}^k a_i=n$,
\begin{equation*}
\prod_{i=1}^k a_i!\sim(2\pi)^{k/2} \Big(\frac nk\Big)^{k/2}\Big(\frac n{ke}\Big)^n
\prod_{i=1}^k(1+h_i)^{a_i},
\end{equation*}
which implies
\begin{equation*}
\binom{n}{a_1,\ldots,a_k}=\frac{n!}{\prod_{i=1}^k a_i!}
\sim(2\pi)^{1/2-k/2}\frac{n^{1/2}}{(n/k)^{k/2}}\frac{k^n}{e^u},
\end{equation*}
where
\begin{equation*}
u:=\sum_{i=1}^k a_i\ln(1+h_i)=\frac nk\sum_{i=1}^k (1+h_i)\ln(1+h_i),
\end{equation*}
by (2).
Note also that (i) $(1+h)\ln(1+h)=h+h^2/2+O(|h|^3)$ as $h\to0$, (ii) $\sum_{i=1}^k h_i=0$ (by the definition of $h_i$ in (1) and the condition $\sum_{i=1}^k a_i=n$), and (iii) $h_i=o(n^{-1/3})$. It follows that
\begin{equation*}
u=\frac n{2k}\,\sum_{i=1}^k h_i^2+o(1)=\frac k{2n}\,\sum_{i=1}^k(a_i-n/k)^2+o(1).
\end{equation*}
Thus,
\begin{equation*}
\binom{n}{a_1,\ldots,a_k}
\sim(2\pi n)^{1/2-k/2}k^{n+k/2}\exp\Big\{-\frac k{2n}\,\sum_{i=1}^k(a_i-n/k)^2\Big\}.
\end{equation*}
In particular, when $k=2$, we get the Wikipedia result quoted in the OP: \begin{equation*} \binom na\sim\frac{2^n}{\sqrt{\pi n/2 }} e^{-2(a-n/2)^2/n} \end{equation*} if $n\to\infty$ and $a=\frac n2+o(n^{2/3})$.
The analogous estimate in the range where your formula is valid (that is, $a_i=\frac nk+o(n^{1/2})$ - note Wikipedia claims the binomial case is valid for $a_i=\frac n2+o(n^{2/3})$, but I wasn't able to reproduce that estimate) is: $$ \binom{n}{a_1\ a_2\ \ldots\ a_k} \sim \frac{k^n}{(2\pi n)^{(k-1)/2}}\exp\left(-\frac kn\sum_{i=1}^k b_i^2\right), $$ where $b_i=a_i-\frac nk$, that is the difference of $a_i$ from its central value.
This can be obtained by using Stirling's formula, and estimates on quantities of the form $(1+\frac an)^n$.
Up to Theorem 5 in this article by Raygorodskiy, $${\lfloor na_1 \rfloor +\lfloor na_2 \rfloor\,+ \dots +\lfloor na_k \rfloor\ \choose \lfloor na_1 \rfloor ,\,\lfloor na_2 \rfloor , \, \dots , \lfloor na_k \rfloor\,} =\left(\frac 1 {a_1^{a_1}a_2^{a_2}\cdots a_k^{a_k}}+o(1)\right)^n$$ as $n\to \infty$, assuming $a_j>0,\ j=1\dots k,$ and $a_1+a_2+\cdots+a_k=1.$
