Well, if the metric $g$ is  Kähler and real-analytic, then having the exponential map from even one point $p$ be holomorphic makes it flat.  (Probably the assumption of real-analytic is not necessary, but I haven't thought about that, and the following argument that uses it is very easy.)

It's enough to show that a Kähler metric on a neighborhood of $0\in\mathbb{C}^n$ with the property that the linear holomorphic coordinates are Gauss normal coordinates is necessarily flat.  To see this, suppose that $z^i$ are linear coordinates on $\mathbb{C}^n$ such that the given metric $g$ agrees with the standard one at $0$.  Then let $f$ be a Kähler potential for $g$, so that
$$
g=\frac{\partial^2f}{\partial z^i\ \partial\overline{z}^j}\ dz^i d\overline{z}^j.
$$
The condition that the $z^i$ be Gauss normal coordinates centered on $0$ is just that
$$
z^i\ \overline{z}^j\ \frac{\partial^2f}{\partial z^i\ \partial\overline{z}^j}
= |z^1|^2 + \cdots + |z^n|^2.
$$
However, now assuming that $f$ is real-analytic, expand it in a power series in $z^i$ and $\overline{z}^i$.  Then this last identity clearly shows that all of the terms in $f$ must be zero except the pure $z$-terms, the pure $\overline{z}$-terms, and the terms of bidegree $(1,1)$, which must simply be $|z^1|^2 + \cdots + |z^n|^2$.  Thus, $g$ is the standard metric.