Just to close this off: note that $d=\partial+\bar\partial$ and that $\partial^2=0$ so $\partial\bar\partial=d\bar\partial$, and therefore $\Omega+\partial\bar\partial\varphi=\Omega+d\bar\partial\varphi$ is in the same cohomology class as $\Omega$. Since wedge product of forms descends to the usual product in cohomology, $(\Omega+d\bar\partial\varphi)^n=\Omega^n$ in cohomology, giving the same volume integral over our compact manifold.

Just to close this off: note that $d=\partial+\bar\partial$ and that $\partial^2=0$ so $\partial\bar\partial=d\bar\partial$, and therefore $\Omega+\partial\bar\partial\varphi=\Omega+d\bar\partial\varphi$ is in the same cohomology class as $\Omega$. Since wedge product of forms descends to the usual product in cohomology, $(\Omega+d\bar\partial\varphi)^n=\Omega^n$ in cohomology, giving the same volume integral over our compact manifold. On the other hand, since the Monge-Ampere equation is elliptic, scalar, determined, it is locally solvable, so $\Omega+d\bar\partial\varphi$ can achieve any multiple of any given volume form, locally, by suitable local choice of $\varphi$. So we cannot guaranteed that $\Omega+d\bar\partial\varphi$ is not zero somewhere, if we allow arbitrary choice of $\varphi$. So we can't be sure that this $(\Omega+\partial\bar\partial\varphi)^n$ is actually a volume form, i.e. a nowhere-zero top-degree form with positive integral. That requires more information.