There's a distinctly non-zero chance that my calculations are wrong, but I think it has a lot of $2$-torsion for $G=A_5$, the alternating group of degree $5$. I'm afraid the method I've used is a little indirect, and I haven't extracted an explicit torsion element.

Certainly $\mathbb{Z}G\otimes_{Z(\mathbb{Z}G)}\mathbb{Z}G$ is a finitely generated abelian group, and so it will be free if and only if $\operatorname{Hom}_\mathbb{Z}(\mathbb{Z}G\otimes_{Z(\mathbb{Z}G)}\mathbb{Z}G,\mathbb{F})$ has the same $\mathbb{F}$-dimension for every field $\mathbb{F}$.

But $$\operatorname{Hom}_\mathbb{Z}(\mathbb{Z}G\otimes_{Z(\mathbb{Z}G)}\mathbb{Z}G,\mathbb{F})\cong \operatorname{Hom}_{Z(\mathbb{Z}G)}\left(\mathbb{Z}G,\operatorname{Hom}_\mathbb{Z}(\mathbb{Z}G,\mathbb{F})\right),$$ which is isomorphic to $\operatorname{Hom}_{Z(\mathbb{F}G)}(\mathbb{F}G,\mathbb{F}G),$ the endomorphism algebra of the group algebra $\mathbb{F}G$ as a module over its centre.

For $\mathbb{F}=\mathbb{C}$ this is easy to calculate from the irreducible character degrees. It's the sum of the fourth powers of the degrees, which for $G=A_5$ is $1^4+3^4+3^4+4^4+5^4=1044$.

For $\mathbb{F}$ algebraically closed of characteristic two, $\mathbb{F}A_5$ has a simple non-principal block isomorphic to $M_4(\mathbb{F})$ which contributes $4^4=256$. The centre of the principal block has a fairly simple structure $Z=\mathbb{F}[x,y,z]/(x,y,z)^2$, where $x,y$ and $z$ are in the socle of the principal block, each annihilated by all but one of the conjugacy classes of primitive idempotents, and according to my calculation, as a module for this centre, the principal block is the direct sum of one copy of $Z/(y,z)$, four copies of each of $Z/(x,y)$ and $Z/(x,z)$, and 26 copies of $Z/(x,y,z)$, and the endomorphism algebra is $1258$-dimensional, so that (taking into account the non-principal block) the endomorphism algebra for the whole group algebra as a module for its centre is $1514$-dimensional.

This can't be the best way to do it, but maybe it provides some clues for a more illuminating answer.

By the way, I tried some smaller groups ($A_4$ in characteristic $2$, $D_{2p}$ in characteristic $p$), and for those the dimensions were the same.

There's a distinctly non-zero chance that my calculations are wrong, but I think it has a lot of $2$-torsion for $G=A_5$, the alternating group of degree $5$. I'm afraid the method I've used is a little indirect, and I haven't extracted an explicit torsion element.

Certainly $\mathbb{Z}G\otimes_{Z(\mathbb{Z}G)}\mathbb{Z}G$ is a finitely generated abelian group, and so it will be free if and only if $\operatorname{Hom}_\mathbb{Z}(\mathbb{Z}G\otimes_{Z(\mathbb{Z}G)}\mathbb{Z}G,\mathbb{F})$ has the same $\mathbb{F}$-dimension for every field $\mathbb{F}$.

But $$\operatorname{Hom}_\mathbb{Z}(\mathbb{Z}G\otimes_{Z(\mathbb{Z}G)}\mathbb{Z}G,\mathbb{F})\cong \operatorname{Hom}_{Z(\mathbb{Z}G)}\left(\mathbb{Z}G,\operatorname{Hom}_\mathbb{Z}(\mathbb{Z}G,\mathbb{F})\right),$$ which is isomorphic to $\operatorname{Hom}_{Z(\mathbb{F}G)}(\mathbb{F}G,\mathbb{F}G),$ the endomorphism algebra of the group algebra $\mathbb{F}G$ as a module over its centre.

For $\mathbb{F}=\mathbb{C}$ this is easy to calculate from the irreducible character degrees. It's the sum of the fourth powers of the degrees, which for $G=A_5$ is $1^4+3^4+3^4+4^4+5^4=1044$.

For $\mathbb{F}$ algebraically closed of characteristic two, $\mathbb{F}A_5$ has a simple non-principal block isomorphic to $M_4(\mathbb{F})$ which contributes $4^4=256$. The centre of the principal block has a fairly simple structure $Z=\mathbb{F}[x,y,z]/(x,y,z)^2$, where $x,y$ and $z$ are in the socle of the principal block, each annihilated by all but one of the conjugacy classes of primitive idempotents, and according to my calculation, as a module for this centre, the principal block is the direct sum of one copy of $Z/(y,z)$, four copies of each of $Z/(x,y)$ and $Z/(x,z)$, and $26$ copies of $Z/(x,y,z)$, and the endomorphism algebra is $1258$-dimensional, so that (taking into account the non-principal block) the endomorphism algebra for the whole group algebra as a module for its centre is $1514$-dimensional.

This can't be the best way to do it, but maybe it provides some clues for a more illuminating answer.

By the way, I tried some smaller groups ($A_4$ in characteristic $2$, $D_{2p}$ in characteristic $p$), and for those the dimensions were the same.