Count satisfying assignments of CNF formulas over all possible negation assignments
The quantity $\sum_k|\phi_k|$ can be computed in polynomial time, in fact, in uniform $\mathrm{TC}^0$. By double counting, we have
$$\sum_k|\phi_k|=|\{(a,k):a\models\phi_k\}|=\sum_{a\in\{0,1\}^n}|\{k:a\models\phi_k\}|.$$
Now, if $a\in\{0,1\}^n$ is any assignment, and $C$ a clause in $l$ literals, then there are exactly $2^l-1$ ways how to negate the literals in $C$ so that it is satisfied by $a$: the unique unsatisfied instance is the one where every literal is negated or not opposite to what $a$ says. Moreover, for a CNF $C_1\land\dots\land C_m$, we choose the negations of literals of each clause $C_j$ independently, and the whole formula is satisfied iff each $C_j$ is satisfied, hence irrespective of $a$ and of the structure of the formula,
$$|\{k:a\models\phi_k\}|=\prod_{j=1}^m(2^{l_j}-1),$$
hence
$$\sum_k|\phi_k|=2^n\prod_{j=1}^m(2^{l_j}-1).$$