Consider $M = S^2 \times T^4$. Then we can construct a non-Kähler complex structure as follows. Let $L$ be a line bundle over $\mathbb{P}^1$ such that there are two sections $s_1, s_2 \in H^0 (L)$ without common zeros. Thus, $\sigma = (s_1, s_2)$ is a nowhere-vanishing section of the vector bundle $W = L \oplus L$. The Hamiltonian quaternions act on $\sigma$ giving four holomorphic sections $\sigma_1 = \sigma, \sigma_2 = i \sigma, \sigma_3 = j \sigma, \sigma_4 = k \sigma$. Then the quotient $X$ of $W$ by the $\mathbb{Z}^4$-action, acting fiberwise by translations $v \mapsto v + \sum a_i \sigma_i$, is a compact complex three-fold. By calculating the canonical divisor, $X$ is Kähler if and only if $W$ is trivial.

I am wondering if it is the only construction, i.e., if we endow $M$ with an arbitrary complex structure, then whether the universal covering is a holomorphic vector bundle $W$ over $\mathbb{C}$?