I'm not sure what the standard terminology for the following setup is, so I will make up some (but I'd love any references to what the correct terminology is). Let $N$ be a normal subgroup of a group $G$. The relative weight of $N$ in $G$, denoted $wt_G(N)$ is the minimum number elements of $N$ needed to normally generate $N$ inside of $G$. If $G$ is just equal to $N$, then this is the same as the weight of $N$ (i.e. the minimal size of a normal generating set).
Can the relative weight be arbitrarily large for normal subgroups of the free group $F_n$?
The only way I can imagine getting lower bounds for $wt_G(N)$ would be to subject $G$ to some other group $G_0$ so that $N$ maps to a normal subgroup $N_0$ where we can actually compute $wt_{G_0}(N_0)$ (say $G_0$ is finite or something). Then $wt_{G_0}(N_0) \leq wt_G(N)$.
The wreath product ${\mathbb Z} \wr {\mathbb Z}$ is a well known example of a finitely generated group that is not finitely presentable.
It can be defined by the infinite presentation $$F/N = \langle a, b \mid [b,a^{-i}ba^i]=1\ (i > 0) \rangle,$$ where $F$ is the free group on $a$ and $b$.
Now, for each $k>0$, let $N_k$ be the normal closure in $F$ of the images of elements $\{ [b,a^{-i}ba^i] : 1 \le i \le k \}$. So $N$ is the ascending union of the $N_k$.
I claim that $N_k$ cannot be defined as the normal closure in $F$ of fewer than $k$ elements.
To see this, consider the group $$F/M = \langle a, b \mid [[b,a^{-i}ba^i],b] = [[b,a^{-i}ba^i],a] =1\ (i > 0) \rangle,$$
So $M < N$ with $N/M \le Z(F/M)$, and $N/M$ is freely generated as an abelian group by the elements $[b,a^{-i}ba^i]$ for $i > 0$.
So the subgroup of $N/M$ generated by the elements $\{ [b,a^{-i}ba^i] : 1 \le i \le k \}$ is free abelian of rank $k$, and it cannot be generated by fewer than $k$ elements.
Since $N/M \le Z(F/M)$, all subgroups of $N/M$ are central in $F/M$, so the normal closure in $F/M$ cannot be generated by fewer than $k$ elements.