Let $(M,g,J)$ be a Kahler manifold. This means that $J$ is a (integrable) complex structure, so that $(M,J)$ is a complex manifold, $g$ is a smooth metric such that $J$ is $g$-orthogonal, i.e.
$g(JX,JY) = g(X,Y)$
for any smooth vector fields $X$, $Y$ on $M$ and, moreover, the Kahler form is closed.

Ok, now let us assume that given an arbitrary point $p \in M$, there exist local holomorphic coordinates $z_i$ around $p$ such that

$\omega = \frac{i}{2} \sum_k dz_k \wedge d\bar{z}_k$

with $\omega$ being the Kahler form. By the definition of the Kahler form, we have $g(X,Y) = \omega(X, JY)$, so that in the local coordinates $z_i$, one has therefore

$g = \frac{i}{2} \sum_k dz_k \odot d\bar{z}_k$

Hence $g$ is locally Euclidean, so that the Riemann curvature tensor vanishes identically. I think the OP is more interested in the converse though.

Conversely, suppose $(M,g,J)$ is a Kahler manifold with Riemann curvature vanishing identically. Being flat, given an arbitrary point $p \in M$, the local holonomy group $Hol_0(g,p)$ is trivial. Thus starting with a basis $\alpha_1,\cdots,\alpha_m$ of the space of $1$-forms at $p$ of type $(1,0)$ (with $m$ being the complex dimension of $M$), one can locally extend the $\alpha_i$ using parallel transport locally to some parallel $\alpha_i$ defined in some neighborhood of $p$. In particular, these $\alpha_i$ form a local holomorphic coframe, with each $\alpha_i$ being further closed. Then using Dolbeault cohomology, it follows that there exist local holomorphic coordinates $z_i$ such that
$dz_i = \alpha_i$.

Moreover, if at $p$, the $\alpha_i$ were chosen to be unitary (at $p$), it follows that their extensions to a neighborhood are also unitary (and as we have seen, parallel too) so that the metric $g$ can be written locally as:

$ g = \frac{i}{2} \sum_k \alpha_k \odot \bar{\alpha}_k = \frac{i}{2} \sum_k dz_k \odot d\bar{z}_k$

Thus the tensor you are looking for is the Riemann curvature tensor.