Are finite nilpotent groups the only finite groups with abelian Frattini quotient?

This is well known, but here is a quick proof anyway.

Suppose that $G/\Phi(G)$ is nilpotent, and let $P \in {\rm Syl}_p(G)$. Then $P\Phi(G) \unlhd G$ so, by the Frattini Argument, $G = N_G(P)P\Phi(G)= N_G(P)\Phi(G)$. Hence $G = N_G(P)$ by the non-generator property of $\Phi(G)$. So all Sylow subgroups are normal, and $G$ is nilpotent.

Here is my attempt to entertain those who knew the answer.

Question 1. Let $G$ be a finitely generated group such that $G/\Phi(G)$ is Abelian. Is $G$ nilpotent?

The answer is yes if we assume moreover that $G$ is linear (e.g., $G$ is finite). But the answer is no in general, as the the first Girgorchuk group $G_1$ is such that $[G_1, G_1] \subset \Phi(G_1)$. For more examples, see this MO post.

Let $W(G)$ denote the $N$-Frattini subgroup of $G$, that is the intersection of the maximal normal subgroups of $G$ when defined, $G$ otherwise. (See this post for results related to $W(G)$.) Note that $G/W(G)$ is Abelian for any soluble group $G$.

Question 2. Let $G$ be a finitely generated group such that $G/W(G)$ is Abelian. Is $G/\Phi(G)$ Abelian?

The answer is no, because some Sunic group of intermediate growth is a counter-example, see this preprint.

Question 3. Let $G$ be a finite group such that $G/W(G)$ is Abelian. Is $G$ soluble?

The answer is no because of the symetric group $S_n$ with $n \ge 5$. One may try to get a classification though. Possibly useful:

Lemma (Baer). Let $G$ be a group and let $N$ be a normal subgroup of $G$. Then $W(N)$ is a normal subgroup of $G$ and $W(N) \subset W(G)$.