Yes. There is an adjunction between $G$-modules and $\mathbb{Z}G$-bimodules defined as follows: the right adjoint $U: \mathbb{Z}G-{\rm BiMod} \to G-{\rm Mod}$ sends a $\mathbb{Z}G$-bimodule $M$ to the $G$-module $M$ with the action given by $(g,m) \mapsto gmg^{-1}$, and the left adjoint $L: G-{\rm Mod} \to \mathbb{Z}G-{\rm BiMod}$ sends a $G$-module $M$ to the $\mathbb{Z}G$-bimodule $L(M) := M \otimes \mathbb{Z}G = \oplus_{g \in G}M\left<g\right>$ where the right action is given by $(\sum_g m_g\left<g\right>,h) \mapsto \sum_g m_g\left<gh\right>$ and the left action is given by $(h,\sum_g m_g\left<g\right>) \mapsto \sum_gh(m_g)\left<hg\right>$. It then follows from general categorical considerations that there is a natural isomorphism
$$ {\rm HH}^n(\mathbb{Z}G) = {\rm Ext}_{\mathbb{Z} G-{\rm BiMod}}(\mathbb{Z}G,\mathbb{Z}G) = {\rm Ext}_{\mathbb{Z} G-{\rm BiMod}}(L(\mathbb{Z}),\mathbb{Z}G) \cong {\rm Ext}_{G-{\rm Mod}}(\mathbb{Z},U(\mathbb{Z}G)) = {\rm H}^n(G,M) $$
where $M$ is the $G$-module $\mathbb{Z}G$ equipped with the natural conjugation action.