It’s not in $AC^0$, by reduction from parity, which is not in $AC^0$.

Let $x$ be a parity input of length $n$.

For each bit $x_i$, we build an $n times n$ matrix $A_i$, s.t. $A_i$ is the identity matrix $I$ if $x_i$ is $0$. Otherwise, it’s a matrix I’ll call $P_{12}$ which permutes first $2$ elements (i.e. it’s almost $I$, but $a_{11}=a_{22}=0$ and $a_{12}=a_{21}=1$).

The result of the multiplication of these matrices is either $I$ or $P_{12}$.

And finding the parity of $x$ is equivalent to checking whether the result is $I$ or $P_{12}$.

These matrices can be built in depth $1$.

If we can multiply the matrices in $O(1)$ depth, then we can check parity in $O(1)$ depth, which is impossible.