Yes, there is a Hodge decomposition for elliptic complexes on compact oriented Riemannian manifolds.
$L^2$-version. Let $(M,g)$ be a compact oriented Riemannian manifold and let $$ 0 \to \Gamma(E_0) \to \Gamma(E_1) \to \dots \to \Gamma(E_m) \to 0 $$ be an elliptic differential complex with the arrows given by the differential operators $D_0,\dotsc,D_{m-1}$, where each $E_i$ is equipped with a metric and a compatible connection. Thus we have formal adjoints and can form the Laplacians: $$ \Delta_i := D_i^* \circ D_i + D_{i-1} \circ D_{i-1}^*: \Gamma(E_i) \to \Gamma(E_i). $$ Then we get the Hodge decomposition $$ L^2(E_i) = \ker(\Delta_i) \oplus \operatorname{ran}(D_{i-1}) \oplus \operatorname{ran}(D_i^*). $$
I learnt this from our own Liviu Nicolaescu's Lectures on the Geometry of Manifolds. More precisely, this is Chapter 10.4.3 in the notes (my version is from 9. Sep 2018).
EDIT: As per OP's comment, I am adding the smooth version I know of.
$C^{\infty}$-version. Consider the total vector bundle $E := \bigoplus_{i=0}^{m} E_i$ and the Laplacian $\Delta = D \circ D^* + D^* \circ D$, i.e. $\Delta$ is just a tuple of Laplacians $\Delta_i: \Gamma(E_i) \to \Gamma(E_i)$. The following statements are equivalent:
(1) $\Delta$ is Fredholm and $\Gamma(E) = \ker(\Delta) + \operatorname{ran}(\Delta)$ (note, this is just a sum, not a direct sum);
(2) The complex itself is Fredholm and satisfies the Hodge decomposition: $$ \Gamma(E) = \ker(\Delta) \oplus \operatorname{ran}(D) \oplus \operatorname{ran}(D^*) $$ This can be found in Chapter 1 of van den Ban and Crainic's lectures on Analysis on Manifolds, more precisely exercise 1.3.15, if you don't mind following the exercises and working it out from the preceding theory, but I don't know a more self-contained source.