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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.

On the other hand, coming back to complex-valued differential forms, if you merely suppose your manifold to be hermitian than the relation between the three Laplacians is a little bit more complicated. You have to introduce the torsion operator $$\tau=[\Lambda_\omega,\partial\omega]$$ which is of type $(1,0)$ and order zero (observe that if $\omega$ is Kähler then $\partial\omega=0$). With these notations, you have $$\Delta''=\Delta'+[\partial,\tau^*]-[\bar\partial,\bar\tau^*]$$ and $$\Delta=\Delta'+\Delta''-[\partial,\bar\tau^*]-[\bar\partial,\tau^*],$$ so that $\Delta'$, $\Delta''$ and $\frac 12\Delta$ no longer coincide but they differ by linear differential operators of order $1$ only.

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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.