Complexity of #PP2DNF where we also count on the number of clauses
The problem is #P-hard, as we prove in a recent preprint: https://arxiv.org/abs/1908.07093

The way the preprint is phrased is closer to the equivalent formulation of the problem given at the end of my question: you have a bipartite graph $(X \cup Y, E)$ and you want to know how many triples $(X', Y', E')$ of subsets $X' \subseteq X$, $Y' \subseteq Y$, $E' \subseteq E$ there are such that we keep two vertices in $X'$ and $Y'$ connected by an edge in $E'$, i.e., $E' \cap (X' \times Y')$ is non-empty. This can again be equivalently reformulated in database or first-order logic terminology, which is what the preprint uses: you have a structure consisting of facts $R(a_1), \ldots, R(a_n)$, $T(b_1), \ldots, T(b_m)$, and $S(a_{i_1}, b_{j_1}), \ldots, S(a_{i_k}, b_{j_k})$ coding the bipartite graph (without multi-edges, i.e., $(a_{i_p}, b_{j_p}) \neq (a_{i_q}, b_{j_q})$ for all $p \neq q$), and you want to know how many subsets of these sets of facts satisfy the conjunctive query $\exists x \, y ~ R(x) \land S(x, y) \land T(y)$.

In the terminology of the preprint, this query is a self-join-free non-hierarchical conjunctive query (indeed, it is the simplest such example of a query, which we call $Q_1$), and hardness of this problem (the so-called model counting problem) is shown in the preprint: this is the simplest case of Theorem 4.2, with $r=1, s=1, t=1$.