If $(X,\omega)$ is Kähler, then it is always true that $$ \Delta'=\Delta''=\frac 12\Delta, $$ where these three Laplacians are with respect, in order, to $\partial$, $\bar\partial$ and $d$.
This is valid when they act on any space of complex-valued differential forms.
More generally, you can look to differential forms with values in a hermitian vector bundle $E\to X$. In this case, take $D_E$ to be the (unique) Chern connection of $E$ and let $D_E=D'_E+D''_E$ its decomposition in the $(1,0)$ and $(0,1)$ part (then, by definition $D''_E=\bar\partial$).
In this case, you can again compare $\Delta'_E$ and $\Delta''_E$. They no longer coincide, but differ by a order zero operator which is expressed in terms of the curvature $\Theta(E)=D^2_E$ and the (formal) adjoint $\Lambda_\omega$ of the operator $L_\omega=\omega\wedge\bullet$ of wedge product with $\omega$. The relation is $$ \Delta''_E=\Delta'_E+[i\Theta(E),\Lambda_\omega], $$ where $[\bullet,\bullet]$ is the (graded) commutator.