@JeremyRickard That's great. That is indeed a better explanation! Thanks for all your help.

@JeremyRickard I think you have disproved it. We know that $\mathbb{F}G^{op}\otimes_{Z(\mathbb{F}G)}\mathbb{F}G$ and $\operatorname{End}_{Z(\mathbb{F}G)}(\mathbb{F}G)$ have the same dimension and so surjectivity is equivalent to injectivity. The torsion elements you've proved exist must act by zero via the map $$\mathbb{Z}G^{op}\otimes_{Z(\mathbb{Z}G)}\mathbb{Z}G\to\operatorname{Hom}_{Z(\mathbb{Z}G)}(\mathbb{Z}G)$$ but they are not zero in $\mathbb{F}G^{op}\otimes_{Z(\mathbb{F}G)}\mathbb{F}G$. Therefore we don't have injectivity.

@Jeremy Rickard Thanks Jeremy. That's not the answer I wanted but it seems to be correct! I am acually interested in $$\operatorname{Hom}_{Z(\mathbb{F}G)}\left(\mathbb{F}G,\mathbb{F}G\right)$$ and wanted it to be generated by left and right multiplication by $\mathbb{F}G$. I think you've essentially disproved this. Thanks for stopping me wasting my time!

@EhudMeir It is also not necessarily projective. Again, Jeremy's example works as a counterexample. If $\mathbb{Z}G$ were projective as a $Z(\mathbb{Z}G)$-module then $\mathbb{F}_2G$ would be projective as a $Z(\mathbb{F}_2G)$-module. However, since $G$ is a $2$-group $Z(\mathbb{F}_2G)$ is a local ring and so any finite dimensional projective $Z(\mathbb{F}_2G)$-module is a direct sum of copies of $Z(\mathbb{F}_2G)$. Now see Jeremy's comment for a contradiction.

Yes. It seems obvious that there is no torsion but I can't prove it.