Assuming $F$ non-abelian, there is no embedding of $F/\mathcal{D}^nF$ for $n\ge 3$ into a group of matrices over any field. (Here I write $\mathcal{D}G=[G,G]$, so $F_n=\mathcal{D}^nG$ in your notation.) One reason is that any solvable group of matrices is nilpotent-by-abelian-by-finite.
But $F/\mathcal{D}^nF$ is not nipotent-by-abelian-by-finite when $n\ge 3$. This is well-known, but here's a proof. Clearly it suffices to do the case when $F$ is free of rank 2. Indeed assume, by contradiction, otherwise. Then it follows that for some $k_0$, $F/\mathcal{D}^nF$ has a subgroup of index $k_0$ that is nilpotent-by-abelian; hence $F/\mathcal{D}^3F$ (being a quotient of $F/\mathcal{D}^nF$) has the same property. In turn, it follows that every 2-generated $3$-solvable group has a subgroup of index $\le k_0$ that is nilpotent-by-abelian.
Now in order to get a contradiction, consider three primes $p,q,r$, all $>k_0$, such that $q$ divides $r-1$, and fix a nontrivial action of the cyclic group $C_q$ on $C_r$. Consider the wreath product $G=(C_r\rtimes C_q)\rtimes C_p$. Let's show it's 2-generated. It clearly 3-generated: let $s,t,u$ be generators of $C_p$, $C_q$, $C_r$. If $g\in C_r\rtimes C_q$ and $k$ is integer, write $g_k=s^kgs^{-k}$ (so $g=g_0$). Then $G=\{s,t_0,u_0\}$. Now observe that $G=\{s,t_0u_1\}$ (use that $t_0$ and $u_1$- commute and have coprime orders), so $G$ is 2-generated. Also, the derived subgroup of $G$ contains $C_r^p$ (since $C_r$ is in the derived subgroup of $C_r\rtimes C_q$), and contains the element $t_Ot_1^{-1}$, which is a commutator of $s$ and $t_0$. In particular, the derived subgroup $\mathcal{D}G$ of $G$ contains an element of order $r$, say $u_0$ and an element of order $q$, say $t_0t_1^{-1}$, that do not commute; hence $\mathcal{D}G$ is not nilpotent; in other words, $G$ is not nilpotent-by-abelian.
Now turning back to the contradictory hypothesis, $G$ should have a nilpotent-by-abelian subgroup $H$ of index $\le k_0$. Since $G$ is not nilpotent-by-abelian, we have $H\neq G$. On the other hand, because $p,q,r>k_0$, $G$ has no proper index subgroup of index $\le k_0$. We get a contradiction, and hence $F/\mathcal{D}^nF$ is not nilpotent-by-abelian-by-finite (and nor is its profinite completion, by the way) and in particular $F/\mathcal{D}^nF$ is not isomorphic to a group of matrices over any field (or product of fields).
