Log-space reduction from Parity-L to CNOT circuits?
Let us start with the $\oplus L$-complete problem of counting mod $2$ the number of paths of length $n$ from vertex $s$ to vertex $t$ in a directed graph $G=(V,E)$. We apply a couple of logspace reductions as follows.

Let $G&#39;=(V&#39;,E&#39;)$ be the graph such that $V&#39;=V\times\{0,\dots,n\}$ and $E&#39;=\{((u,i),(v,i+1):i&lt;n,(u,v)\in E\}\cup\{(w,w):w\in V&#39;\}$ (i.e., we take $n+1$ copies of $G$’s vertices, make edges go from the $i$th copy to the $(i+1)$th copy according to $G$’s edges, and add all self-loops). Then the original problem is equivalent to counting paths of length $n$ from $s&#39;=(s,0)$ to $t&#39;=(t,n)$ in $G&#39;$.

Moreover, $G&#39;$ is acyclic, and we can explicitly define an enumeration $V&#39;=\{w_k:k\le m\}$ such that all edges in $G&#39;$ apart from the self-loops go from $w_k$ to $w_l$ for some $k&lt;l$. Without loss of generality, $w_0=s&#39;$ and $w_m=t&#39;$. Let $M$ be the adjacency matrix of $G&#39;$ wrt the given enumeration. Then $M$ is an upper triangular integer matrix with $1$ on the diagonal, and the number of paths of length $n$ from $s&#39;$ to $t&#39;$ equals the top right element of $M^n$.

It is easy to see that
$$M=\prod_{j=m}^1\prod_{i=0}^{j-1}E_{i,j}(M_{i,j}),$$
where $E_{i,j}(a)$ is the elementary matrix whose only nondiagonal entry is $a$ in row $i$ and column $j$. In this way, we reduced the original problem to computing the top right element of a product of elementary matrices. In the $\oplus L$ case, the computation is modulo $2$, i.e., we consider the matrices over $\mathbb F_2$. (In this case, the elementary matrices can be only $E_{i,j}(0)=I$, which we can ignore, and $E_{i,j}(1)$, which can be simulated by a single CNOT gate, as mentioned in the question.) If we consider them as integer matrices, we get a $\#L$-complete problem, and if we consider them modulo $k$, we get a $\mathrm{Mod}_kL$-complete problem.