From the discussion in the comments two possible solutions arised, here is the first:
Since the group is nilpotent, upper central series terminates. Then for some $n$, $Z_n(G) = G$ and there must be a smallest $i$ such that $N \cap Z_i(G) = \langle e \rangle$ but $N \cap Z_{i+1}(G) \neq \langle e \rangle$. The goal is to show that $N \cap Z_{i+1}(G) \subseteq Z_1(G) = Z(G)$.
I will denote $Z_i(G),Z_{i+1}(G)$ simply by $Z_i, Z_{i+1}$.
We notice that the subgroup $[G,N]$ generated by the elements $\{ gng^{-1}n^{-1}\mid g \in G, n \in N\}$ is contained in the normal subgroup $N$. This is because for every generator $gng^{-1}n^{-1}$ of $[G,N]$, $gng^{-1}n^{-1} = (gng^{-1})n^{-1} \in N$. So in conclusion $[G,N] \subseteq N$.
Further we notice that since $Z(G/Z_i) = Z_{i+1}/Z_i$, then for any $c \in Z_{i+1}, g \in G$ we have that $Z_igc = Z_icg$, which implies $Z_igcg^{-1}c^{-1} = Z_i$ and $gcg^{-1}c^{-1} \in Z_i$ for all $g \in G, c \in Z_{i+1}$. These are exactly the generators of the group $[G, Z_{i+1}]$, therefore $[G, Z_{i+1}] \subseteq Z_i$
Now we look again at the nontrivial group $N \cap Z_{i+1}$. We established earlier that $[G,N] \subseteq N$ and $[G, Z_{i+1}] \subseteq Z_i$. This gives $[G, N \cap Z_{i+1}] \subseteq [G,N] \cap [G,Z_{i+1}] \subseteq N \cap Z_i = \langle e\rangle$. So actually the subgroup $[G, N \cap Z_{i+1}]$ is trivial, which happens when $N \cap Z_{i+1} \subseteq Z_1 = Z(G)$. This gives that $N \cap Z(G) \neq \langle e \rangle$.
And a perhaps a bit simpler one, relying on an alternative characterisation of nilpotent groups:
By an exercise in my book (Hungerford's Algebra, chapter 2, section 7, exercise 4), a group is nilpotent if and only if the $\gamma_m(G) = \langle e \rangle$ for some $m$, where $\gamma_1(G)=G,\gamma_2(G)=[G,G]$ and $\gamma_i(G)=[\gamma_{i−1}(G),G]$.
Given this, we define a sequence $N_1(G) = N, N_2(G) = [N,G]$ and $N_i(G) = [N_{i-1}(G),G]$, where $N$ is any proper normal subgroup of $G$.
Obviously for any $i$, $N_i(G) \subset N$ by normality of $N$.
It is also clear that $N_1(G) = N \subset G = \gamma_1(G)$. Assume inductively that $N_i(G) \subset \gamma_i(G)$. By definition, $N_{i+1}(G) = [N_{i}(G),G]$ and $\gamma_{i+1}(G)=[\gamma_{i}(G),G]$. It is clear then that $N_{i+1}(G) \subset \gamma_{i+1}(G)$.
But since $G$ is nilpotent, for some $m$, $N_m(G) \subset \gamma_m(G) = \langle e \rangle$ So in particular we get that $[N_{m-1}(G), G] = \langle e \rangle$. This can only be if $N_{m-1}(G) \subset Z(G)$, and since $N_{m-1}(G) \subset N$, we have that $N \cap Z(G) \neq \langle e \rangle$.
This is Problem $2B.1$ of Isaacs' Finite Group Theory. You just need to combine elements from the proofs of Theorem $2.8$ and $2.12$ of the same book into one statement.
Induct on $|G|$, $|G|=1$ being trivial. Let $K$ be a proper subgroup of $G$ containing $H$. Then for each $k \in K$ either $\langle H,H^k \rangle$ is nilpotent or $HH^k = H^kH$. So by the induction hypothesis $H \lhd \lhd K$. Suppose $H$ is not subnormal in $G$. Then the Zipper Lemma guarantees the existence of a unique maximal subgroup $M$ with $H \subseteq M$. Let $g \in G$, there are two cases.
$(1)$ $\langle H,H^g \rangle$ is nilpotent. Since every subgroup of a finite nilpotent group is subnormal, $\langle H, H^g \rangle$ is proper
in $G$. Therefore $\langle H, H^g \rangle \subseteq M$.
$(2)$ $HH^g=H^gH$. Then $HH^g$ is a subgroup of $G$. But, $H$ is proper and hence $HH^g$ is proper. Thus $HH^g \subseteq M$.
In either case, $H^g \subseteq M$. Thus the normal closure $H^G \subseteq M$, whence $H^G$ is proper in $G$. But then we have $H \lhd \lhd H^G \lhd G$, contradicting $H$ was assumed to be not subnormal in $G$.
Best Answer
I am assuming that by $F(G)$ you mean the subgroup generated by all normal nilpotent subgroups of $G$.
Every element of $F(G)$ can be written as a product of some generators, so any $x \in F(G)$ is contained in $\langle N_1, \ldots, N_k \rangle$ for some nilpotent normal subgroups $N_1$, $\ldots$, $N_k \trianglelefteq G$.
Thus any finitely generated subgroup of $F(G)$ is contained in $\langle N_1, \ldots, N_k \rangle$ for some nilpotent normal subgroups $N_1$, $\ldots$, $N_k \trianglelefteq G$. So $F(G)$ is locally nilpotent.