Let $G$ be (finite) matabelian group; define $W(G):=C_G(C_G(G'))$; $G'$ is derived subgroup of $G$.
If $\mathcal{F}$ is the collection of all maximal abelian normal subgroups of $G$ which contain $G'$, then for any $A\in\mathcal{F}$, it is easy to show that $C_G(A)=A$. Since $G'\le A$, so $C_G(G')\ge C_G(A)=A$, so $C_G(C_G(G'))\le C_G(A)=A$ for all $A\in\mathcal{F}$; hence $$W(G)=C_G(C_G(G'))\le \bigcap_{A\in\mathcal{F}} A. $$
Q.1 Is there non-nilpotent metabelian group where $W(G)$ is proper subgroup of intersection of all maximal abelian normal subgroups in $G$?
Q.2 In metabelian groups $G$, what properties about $W(G)$ are known? Is there special name to this subgroup?
In the case of nilpotent groups, one can easily obtain example of this nature: take a $p$-group $H$ in which $H'$ is proper in $Z(H)$. Then $W(H)=Z(H)$; so to push $W(H)$ properly in intersection of all maximal abelian normal subgroups, we may look for a $p$-group in which maximal abelian normal subgroup is unique. I think, the construction of such $p$-group is not difficult; however, I was unable to find an example of non-nilpotent group.
Edit (after comment by Holt): I think, question (1) is not meaningful with following computation:
Let $G$ be (finite) metabelian; let $A_1,\ldots, A_l$ be all the maximal abelian normal subgroups containing $G'$. (So, $G/A_i$ is abelian, and it can be seen that $C_G(A_i)=A_i$.)
Suppose $g\in C_G(G')$. Note that $G'$ is abelian (since $G$ is metabelian). Then $\langle g,G'\rangle$ is abelian normal subgroup of $G$ (since it contains $G'$), so $\langle g,G'\rangle\subseteq A_i$ for some $i$, hence $g\in A_i$ for some $i$. This means, $C_G(G')\subseteq \cup_i A_i$. Then taking centralizer in $G$, we get $C_G(C_G(G'))\supseteq \cap_i C_G(A_i)=\cap_i A_i$. Thus $W(G)\supseteq \cap A_i$. The reverse inclusion is shown before Q.1. Thus, for finite metabelian equality holds in equation before Q.1. (I hope this is correct. Comments will be helpful.)