If we have a central extension of group schemes $1\rightarrow B \rightarrow C\rightarrow
A\rightarrow1$ with $A$ abelian, then we get a commutator mapping
$\Lambda^2A\rightarrow B$ (of sheaves as $\Lambda^2A$ in general is not a group
scheme) and the extension is abelian precisely when this map is zero. Hence for
an non-abelian extension to exist there must be a non-zero map
$\Lambda^2A\rightarrow B$. Let us now assume that $A=\mu_n$ and consider first
the case when $n=p$, the characteristic of the field $k$ (which we may assume is
algebraically closed). A non-zero map $\Lambda^2A\rightarrow B$ would give a
non-zero map $A\rightarrow\mathrm{Hom}(A,B)$, where the right hand side is the
sheaf of group homomorphisms. As the Frobenius map is zero on $\mu_p$ we may
replace $B$ by its Frobenius kernel so we may assume that $B$ is either $\mu_p$
or the Cartier dual of $\alpha_{p^m}$. Now, as sheaves $\mathrm{Hom}(A,B)$ is
isomorphic to $\mathrm{Hom}(D(B),D(A))$, where $D(-)$ denotes the Cartier
dual. However $D(\mu_p)=\mathbb Z/p$ so when $A=\mu_p$ we get that
$\mathrm{Hom}(D(A),D(B))=\mathbb Z/p$ and there is only the zero map from
$A=\mu_p$ into it. In the other case $D(A)=\alpha_{p^m}$ and
$\mathrm{Hom}(\alpha_{p^m},\mathbb Z/p)$ is zero. If instead $n=p^k$, the
argument is the same. The case when $n=\ell^k$ is even simpler so in all cases
all possible commutator maps are zero and the extension is commutative.

(When $A=B=\mathbb G_a$ then there are candidates for commutator maps and in
fact $(a,b)(a',b')=(a+a',b+b'+a^pa')$ gives a non-commutative central extension
which I imagine is the fake Heisenberg group.)