I do not like complex numbers and can make a mistake easily...
Let $L_p$ be a complex line in a tangent space $T_pX$. It is easy to see that $\exp_p$ gives an isometric embedding $L_p\hookrightarrow X$ which is also star-shaped with center at $p$; set $L=\exp_p(L_p)$
Take any other point $q\in L$, and let $L_q\subset T_q$ be the tangent subspace to $L$. Note that the maps $\exp_p$ and $\exp_q$ coinside (up to a shift) on the geodesic $(pq)$. [Here I use that if two holomorphic maps coincide on the real line then they coincide in the complex plane.] It follows that $L=\exp_q(L_q)$. Therefore $L$ is totally geodesic.
In other words: For any complex sectional direction in $X$, there is a tangent totally geodesic surface which is isometric to complex plane.
Does not it imply that
In particular, the sectional curvature in all complex sectional directions is zero and therefore the curvature of $X$ is isometric identically zero; the later stated in Kobayashi--Nomizu, Foundations of differential geometry, Volume 2 IX, Prop. 7.1. (thanks to $\mathbb C^n$?RdN).
