Treat $C^n$ as a complex manifold, and consider the standard Hermitian metric $h = \frac{1}{2} \sum_{i=1}^n dz_i d\bar{z}_i$, so that the corresponding J-invariant metric is $g = \sum dx_i^2+dy_i^2$, where $z=x+iy$. Define $\Delta = dd^{*}+d^{*}d : \Omega^{p,q} \rightarrow \Omega^{p,q}$, where $d^* = -*d*$, where $*$ is the Hodge operator.
I computed $\Delta(f) = -2 \displaystyle{\sum_{i=1}^n \frac{\partial{}}{\partial{\bar{z_i}}}\frac{\partial{f}}{\partial{z_i}}}$, but I was expecting the factor to be $-4$ at the front.
Here is my computation: Let $v = dx_1 \wedge dy_1 \wedge ... \wedge dx_n \wedge dy_n = C dz_1 \wedge d\bar{z_1} \wedge ...$ be the volume form, for some constant $C$. I believe $C = (i/2)^n$. Then, since $(d\bar{z_i}, d\bar{z_i}) = 2$ using the induced metric on the dual space, we get that $*dz_i = -2C dz_1 \wedge d\bar{z_1} \wedge ... \wedge dz_i \wedge dz_{i+1}\wedge d\bar{z_{i+1}} ...$ so
$\displaystyle{-*d*df = -*d* \frac{\partial{f}}{\partial{z_i}} dz_i = 2C *d \frac{\partial{f}}{\partial{\bar{z_i}}} d\bar{z_i} \wedge dz_1 \wedge d\bar{z_1} \wedge ... \wedge dz_i \wedge dz_{i+1} = -2 \frac{\partial{}}{\partial{\bar{z_i}}}\frac{\partial{f}}{\partial{z_i}} * v}$, and $*v = 1$.