Almost the exact same question was asked here, but nobody proved or cited its #P-completeness! I found this question because I proved it is #P-complete (proof below), and the proof was trivial, but I couldn't find it anywhere on the web or in the discussion on that question! (One person commented "I suspect it is #P-complete".) So either my proof is wrong, which you can judge for yourself, or the result is novel, which I would think is highly unlikely. Which is it?
Proof:
Consider the set of regular expressions formed by the union and concatenation of symbols from $(\Sigma\cup{.})$, where $\Sigma$ is an alphabet and $.$ is a symbol for any $l\in\Sigma$. I will show that #SAT reduces in $O(mn)$ time to counting the number of strings that satisfy such a regex, where $m$ is the number of clauses and $n$ is the number of variables in the given formula $\varphi$.
The reduction is very simple. First, reduce #SAT to #UNSAT by noting that if $\varphi$ has $n$ variables, then $\#\text{SAT}(\varphi) = 2^n - \#\text{UNSAT}(\varphi)$. So now we have to count the number of falsifying assignments of the variables of $\varphi$.
To falsify $\varphi$, it suffices to falsify just one clause. We therefore want to find $$\left|\bigcup \limits_{C\in\varphi}F(C)\right|$$ where $F(C)$ is the set of assignments of literals that falsify $C$.
We thus write a regex representing the language of falsifying assignments for each clause. Using $\Sigma = \{0,1\}$, and letting $r_i$ be the $i^{th}$ symbol of a regex $r$, the regex for the language of assignments that falsify a clause $C$ is $$ r_i = \begin{cases} 0 & \text{if } x_i\in C \\ 1 & \text{if } \overline{x_i}\in C \\ . & \text{otherwise} \end{cases} $$
We then take the union of the regexes for each clause $C\in\varphi$. For example, if $\varphi = (x_1\vee \overline{x_2}\vee x_3) \wedge (x_1\vee x_3\vee \overline{x_4})\wedge (\overline{x_1}\vee \overline{x_4})$ we would have $r = $010.|0.01|1..1, where | denotes union.
The construction yields a union of $m$ strings, each of $n$ symbols. Thus the reduction takes $O(mn)$ time and the size of the regex-counting instance is polynomial in the size of the formula. Further, any regular expression that uses only concatenation, dot, and union represents a finite regular language (specifically, there is no Kleene star). So since we can solve #SAT by counting strings that satisfy such a regex, we can say that counting the words in a finite language is #P-complete.
