At least computing $F_k(x)$ turned out not to be that hard after all. Slightly more generally, consider
$$ \sum_{d|k} Z_d(x) F_{n/d}(x^d) = L_n(x), $$ with $Z_1(x)=1$.
Then we have $$ F_n(x)=\sum_{1=d_0|d_1|\dots|d_k|n}L_{n/d_k}(x^{d_k}) (-1)^k\prod_{i=0}^{k-1} Z_{d_{i+1}/d_i}(x^{d_i}), $$
where in the sum $d_0 < d_1 < \dots < d_k \leq n$. In other words, we are summing over all chains starting at $1$, below $n$. The formula is easily shown by induction.