You can just prove it yourself directly in local holomorphic coordinates. Indeed, the $\overline{\partial}$ Laplacian on functions is equal to $\Delta_{\overline{\partial}}f=g^{i\overline{j}}\partial_i \partial_{\overline{j}}f$. Apply this to $|\partial f|^2=g^{k\overline{\ell}}\partial_k f \partial_{\overline{\ell}}f$ (the gradient of the complex length squared of $f$, \partial f=(df)^{(1,0)}$, which is equals $1/2$ of the usual $|\nabla f|^2$), using if you want local holomorphic normal coordinates for $g$ at a point, and you will immediately get
$$\Delta_{\overline{\partial}}|\partial f|^2=|\nabla_i \nabla_j f|^2+|\nabla_i \nabla_{\overline{j}} f|^2+2\mathrm{Re}\langle \partial f, \partial\Delta_{\overline{\partial}}f\rangle+R^{i\overline{j}}\partial_i f\partial_{\overline{j}}f,$$ where $R^{i\overline{j}}$ is the Ricci curvature of $g$ with the indices raised.
If $g$ is not Kähler, and you define the complex Laplacian by the same formula $g^{i\overline{j}}\partial_i \partial_{\overline{j}}f,$ then a similar result holds, with the Ricci curvature now being one of the several Ricci curvatures of the Chern connection of $g$, and with several new terms involving the torsion of $g$ and its covariant derivative. The calculation is again completely strightforward, using local holomorphic coordinates (not normal anymore!), and using the definitions of covariant derivative and curvature of the Chern connection of $g$.
