That Stein manifolds have all $(p,q), p \geq 0, q \geq 1$ vanishing Dolbeault cohomology groups is more or less standard. I am a little bit confused about the reverse implication: whether the vanishing of all Dolbeault cohomologies ($p \geq 0, q \geq 1$) implies Steinness? I could not find a reference, yet I noticed the following:

In the recent (and quite authoritative) book by Forstnerič ''Stein Manifolds and Holomorphic Mappings'' Theorem 2.4.6 says:

On any Stein manifold $X$ the Dolbeault cohomology groups vanish: $H^{p,q}_{\bar\partial}(X) = 0$ for all $p \geq 0, q \geq 1$.

So the reverse is not even mentioned.

In the older book by Grauert and Remmert ''Theory of Stein spaces'' on page 81 one reads:

The assumptions of Theorem 5 are always satisfied by Stein manifolds. It is not known if there exist non-Stein manifolds of this type (i.e. $H^q(X,\Omega^p)=0, p \geq 0, q \geq 1$)

So I assume this was open at least as of 2004 (?)

On the other hand a recent paper by Chakrabarti and Shaw entitled ''The $L^2$-cohomology of a bounded smooth Stein domain is not necessarily Hausdorff'' (on arXiv, the author's website says that it is to appear in Math. Ann.) begins with:

For each bidegree $(p,q)$, with $p\geq0,q >0$, the Dolbeault Cohomology group $H^{(p,q)}(\Omega)$ of a Stein manifold $\Omega$ in degree $(p,q)$ vanishes, and indeed this property characterizes Stein manifolds among complex manifolds.

The provide references for that fact are the books of Hörmander and Gunning-Rossi (where I couldn't find it).