At least computing $F_k(x)$ turned out not to be that hard after all. Slightly more generally, consider
$$ \sum_{d|k} Z_d(x) F_{n/d}(x^d) = L_n(x). $$
Then it seems (Warning: I didn't prove this yet!) that
$$ F_n(x)=L(n)+\sum_{1< d|n} L_{n/d}(x)\sum_{1=d_0|d_1|\dots|d_{k+1}=d}(-1)^{k+1}\prod_{i=0}^k Z_{d_{i+1}/d_i}(x^{d_i}), $$
where in the second sum all divisors are proper, i.e., $d_i < d_{i+1}$. In other words, we are summing over all chains from $1$ to $d$.