I guess your diffeomorphism is a biholomorphism, or at least an holomorphic map, otherwise there is no hope to say much about it.

Then the best way to look at these things is using differential forms. Indeed, $\imath \partial \bar{\partial} u =\imath \sum_{i,j} \frac{\partial^2 u}{\partial z_i \partial \bar z_j} dz_i \wedge d\bar z_j$. As is well known, as $\varphi$ is holomorphic (not necessarily bijective), $\varphi^* $ commutes with $\partial$ and $\bar \partial$. Therefore $\imath \partial \bar{\partial} u' = \varphi^* \imath \partial \bar{\partial} u$, which you can translate in terms of matrices then, if you want to.

I guess your diffeomorphism is a biholomorphism, or at least an holomorphic map, otherwise there is no hope to say much about it.

Then the best way to look at these things is using differential forms. Indeed, $\imath \partial \bar{\partial} u =\imath \sum_{i,j} \frac{\partial^2 u}{\partial z_i \partial \bar z_j} dz_i \wedge d\bar z_j$. As is well known, as $\varphi$ is holomorphic (not necessarily bijective), $\varphi^* $ commutes with $\partial$ and $\bar \partial$. Therefore $\imath \partial \bar{\partial} u' = \varphi^* \imath \partial \bar{\partial} u$, which you can translate in terms of matrices then, if you want to:

$$Hess(u')_a(\xi)= \underset{j,k,l,m}{\sum} \frac{\partial^2 u(\varphi(a))}{\partial z_l \partial \bar z_m} \frac{\partial \varphi_l(a)}{\partial z_j } \xi_j \overline{\frac{\partial\varphi_m(a)}{\partial z_k } \xi_k} $$ if $\varphi= (\varphi_1, \ldots, \varphi_n)$, and $\xi$ is any tangent vector at $a$.