To get the Dolbeault complex, you need a choice of holomorphic structure on $E$, not just a smooth one. If $\mathcal{E}$ is the locally free sheaf of $\mathcal{O}_M$-modules corresponding to $E$, then $\mathcal{E}\subset \mathcal{A}^0(\mathcal{E})=\mathcal{C}^\infty_M\otimes_{\mathcal{O}_M}\mathcal{E}$ and $\overline{\partial}_E: \mathcal{A}^0(\mathcal{E})\to \mathcal{A}^{0,1}(\mathcal{E})$ is the unique morphism (of sheaves of $\mathbb{C}$ vector spaces) satisfying $$ \overline{\partial}_E(f\sigma) = \overline{\partial}f\otimes \sigma + f\overline{\partial}_E(\sigma),$$ for any smooth function $f$ and $\sigma$ a smooth section of $E$, such that $\left. \overline{\partial}_E\right| _{\mathcal{E}}=0$. The first term in the Leibniz formula involves $\overline{\partial}=d^{0,1}$. You can then extend the Dolbeault operator to $\overline{\partial}_E: \mathcal{A}^{0,p}(\mathcal{E})\to \mathcal{A}^{0,p+1}(\mathcal{E})$, $\overline{\partial}_E^2=0$, by imposing the Leibniz rule with the usual sign. This gives you the Dolbeault resolution $$ 0\to \mathcal{E}\to \mathcal{A}^0(\mathcal{E})\to \mathcal{A}^{0,1}(\mathcal{E})\to\ldots$$ The complex you write is obtained by passing to global sections of $\mathcal{A}^{0,\bullet}(\mathcal{E})$.
You cannot do any of this without the holomorphic structure. Differently put, you need the total space of $E$ to be a complex manifold and the projection $E\to M$ to be holomorphic. You cannot play the same game in the real case, but if you are willing to assume that $E$ carries a flat connection, then you can look at the de Rham resolution of the corresponding local system, as David explains.
ADDENDUM
The requirement that a (smooth) complex vector bundle $V$ admits a holomorphic structure $\overline{\partial}_E$ is non-trivial. It can be phrased as follows:
$V$ admits a holomorphic structure if and only if it admits a connection, $D$, such that $D^{0,1}\circ D^{0,1}=0$, i.e., a connection for which the $(0,2)$ component of the curvature vanishes.