Unfortunately this does not work, because the Hodge $*$ operator is parallel commutes with respect to the Levi-Civita connection(more precicely, it is parallel with respect to the connections induced on natural bundles over $M$). . Indeed, we have $$\nabla (\langle u,v \rangle dV) = \langle \nabla u, v \rangle dV + (-1)^m \langle u, \nabla v \rangle dV = \nabla u \wedge * v + (-1)^m u \wedge *\nabla v$$ because $\nabla dV = 0$ since the metric $g$ is parallel with respect to $\nabla$. The left hand side of this formula is again $$\nabla (u \wedge * v) = \nabla u \wedge *v + (-1)^m u \wedge \nabla(*v),$$ from which we get $\nabla * = * \nabla$. It follows that $* \nabla * = ** \nabla = (-1)^l \nabla$ for some $l$.
There is an expression for $\nabla^*$ in terms of a local orthonormal frame in Werner Ballmann's Lectures on Kahler manifolds (Proposition 1.27, Chapter 1, p. 11) that says that if $(X_1, \ldots, X_n)$ is such a frame, and if $\hat\nabla$ is the dual connection, then $$\nabla^*u = - \sum_j X_j \llcorner \hat\nabla_{X_j} u.$$ I don't know if this is what you're looking for, if not you might have more luck with Bocher-WeitzenbĂ¶ck type identities.
Unfortunately this does not work, because the Hodge $*$ operator is parallel with respect to the Levi-Civita connection (more precicely, it is parallel with respect to the connections induced on natural bundles over $M$). Indeed, we have $$\nabla (\langle u,v \rangle dV) = \langle \nabla u, v \rangle dV + (-1)^m \langle u, \nabla v \rangle dV = \nabla u \wedge * v + (-1)^m u \wedge *\nabla v$$ because $\nabla dV = 0$ since the metric $g$ is parallel with respect to $\nabla$. The left hand side of this formula is again $$\nabla (u \wedge * v) = \nabla u \wedge *v + (-1)^m u \wedge \nabla(*v),$$ from which we get $\nabla * = * \nabla$. It follows that $* \nabla * = ** \nabla = (-1)^l \nabla$ for some $l$.
There is an expression for $\nabla^*$ in terms of a local orthonormal frame in Werner Ballmann's Lectures on Kahler manifolds (Proposition 1.27, Chapter 1, p. 11) that says that if $(X_1, \ldots, X_n)$ is such a frame, and if $\hat\nabla$ is the dual connection, then $$\nabla^*u = - \sum_j X_j \llcorner \hat\nabla_{X_j} u.$$ I don't know if this is what you're looking for, if not you might have more luck with Bocher-WeitzenbĂ¶ck type identities.