Here a possible alternative to the approach outlined in the comments. Let us assume we start with the FL-expansion of $\text{Li}_2(x),\log^2(1-x)$ or $\log(x)\log(1-x)$ (they all are reasonably simple and related to each other via the dilogarithm reflection formula), for instance
$$\text{Li}_2(x) = (\zeta(2)-1)+\sum_{n\geq 1}\frac{2n+1}{n^2(n+1)^2}\,P_n(2x-1).$$
We have
$$ \frac{\text{Li}_2(x)}{x}= \zeta(3)+\sum_{n\geq 1} c_n\,P_n(2x-1) $$
and by Bonnet's recursion formula
$$ x P_n(2x-1) = \frac{n+1}{4n+2}P_{n+1}(2x-1)+\frac{1}{2}P_n(2x-1)+ \frac{n}{4n+2}P_{n-1}(2x-1) $$
so the coefficients of the FL-expansion of $\frac{\text{Li}_2(x)}{x}$ are given by the solution of the recurrence
$$ \frac{n}{4n-2} c_{n-1} + \frac{1}{2} c_n + \frac{n+1}{4n+6}c_{n+1} = \frac{2n+1}{n^2(n+1)^2}\tag{R} $$
with the initial conditions $c_0=\zeta(3)$ and $c_1=\pi^2-6-3\zeta(3)$. My version of Mathematica is not able to directly crack this through $\text{RSolve}$, but we may establish once again the superiority of humans over machines through insights, i.e. our chaotic thought process.
By eye-balling it is not difficult to get
$$ c_n=(2n+1)\int_{0}^{1}\frac{\text{Li}_2(x)}{x}P_n(2x-1)\,dx = (-1)^n(2n+1)\left[\zeta(3)-H_n\frac{\pi^2}{3}+r(n)\right]$$
with $r(n)\in\mathbb{Q}$, and since $c_n\to 0$ we may expect that the structure of $r(n)$ is close to the one of $2 H_n H_n^{(2)}-H_n^{(3)}$. If we let $c_n=(-1)^n (2n+1) d_n$ in $(R)$ we end up with
$$ -n d_{n-1} + (2n+1) d_n - (n+1) d_{n+1} = 2(-1)^n \left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\tag{R'} $$
and by letting $e_n=d_n-d_{n-1}$, then $f_n = ne_n$, we get
$$ n e_n - (n+1) e_{n+1} = 2(-1)^n \left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\tag{R''} $$
$$ f_n - f_{n+1} = 2(-1)^n \left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\tag{R'''} $$
so
$$ f_n = \sum_{k\geq n}2(-1)^k \left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right) $$
$$ e_n = \frac{1}{n}\sum_{k\geq n}2(-1)^k \left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right) $$
$$ d_n = \sum_{m=1}^{n}\frac{1}{m}\sum_{k\geq m}2(-1)^k \left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right) $$
and by dropping the $\zeta(3)$-related part
$$\boxed{ c_n = (-1)^n(2n+1)\sum_{m=1}^{n}\frac{1}{m}\sum_{k\geq m}2(-1)^k \left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right). }$$
A similar approach can be used for finding the FL-expansions of
$\text{Li}_n(x)$, since
$\text{Li}_{n+1}(x)=\int\text{Li}_n(x)\frac{dx}{x}$ and $\int
P_n(x)\,dx=\frac{1}{2n+1}(P_{n+1}-P_n)$. According to my knowledge the
FL-expansion of $\text{Li}_4$ is yet unpublished, but I think it is
not a big deal to directly share this technique with the community (I hope Marco agrees :/)
