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 $i=1,\dots,k$.
Let
\begin{equation}
h_i:=\frac kn\,a_i-1=o(n^{-1/3}).
\end{equation}
By Stirling's formula:
\begin{equation}
m!\sim\sqrt{2\pi m}(m/e)^m
\end{equation}
as $m\to\infty$,

\begin{equation}
\binom{n}{a_1,\ldots,a_k}
\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}
Since $(1+h)\ln(1+h)=h+h^2/2+O(|h|^3)$ as $h\to0$, $\sum_{i=1}^k h_i=0$, and $h_i=o(n^{-1/3})$, we have
\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}
Collecting the pieces, we get
\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 $a=\frac n2+o(n^{2/3})$.