Let $(X,\omega)$ be a Kähler manifold. Call $X$ uniformly Stein with respect to $\omega$ if there exists an exhaustion map $\phi:X\to \mathbb{R}$ such that $i\partial\bar\partial \phi \ge \epsilon\omega$ for some $\epsilon>0$. This implies that $X$ is a Stein manifold.

1. Are there obstructions on $(X,\omega)$ for $X$ to be uniformly Stein with respect to $\omega$ ?

2. Are there examples of Stein manifolds which are not Stein with respect to some Kähler metric ?

I am mostly interested in the case where $X$ is the universal cover of a compact Kähler manifold $X/\Gamma$ and the Kähler metric is the pullback $\pi^*\omega$ of some Kähler metric $\omega$ on $X/\Gamma$. Being uniformly Stein is then intrinsic, in the sense that it does not depend on the choice of the Kähler metric $\omega$ on $X/\Gamma$.

3. Are there examples of compact Kähler manifolds whose universal cover is Stein but not uniformly Stein ?

For universal covers of compact manifolds, it is enough to produce a map $\phi:X\to \mathbb{R}$ such that $i\partial\bar\partial \phi \ge \epsilon\pi^*\omega$, because of the existence of an exhaustion whose Levi form is bounded from below (T. Napier. Convexity properties of coverings of smooth projective varieties. Mathematische Annalen. 1990.)