NB:  I'm combining my previous comments into an answer, because I believe that this is better than leaving them scattered.

As another commenter has pointed out, the skew-symmetric part of the Ricci tensor is the obstruction to there being a $\nabla$-parallel volume form in the first place.  To see this, you use the Bianchi identity: First Bianchi is $R^i_{jkl}+R^i_{klj}+R^i_{ljk}=0$. Set $i=j$ and sum to get $R^i_{ikl}+R^i_{kli}+R^i_{lik}=0$, which becomes $R^i_{ikl}=R^i_{kil}-R^i_{lik}$.  Now $\Omega = \frac12 R^i_{ikl}\ dx^k\wedge dx^l$ is the curvature of the connection induced by $\nabla$ on the top exterior power of the cotangent bundle, and $\frac12(R^i_{kil}-R^i_{lik})dx^k\wedge dx^l$ is the skew-symmetric part of the Ricci tensor.  Thus, the vanishing of the skew-symmetric part of Ricci is equivalent to the flatness of this induced connection on the top exterior power.

If you go ahead and assume that the Ricci curvature is symmetric, so that there is a (local) $\nabla$-parallel volume form, say $\Upsilon$, then the Ricci curvature has the following interpretation:  Let $\exp_p:T_pM\to M$ be the exponential map of $\nabla$ based at $p$.  Then 
$$
\exp^\ast_p(\Upsilon)=(1 - \tfrac13 R_{ij} x^ix^j + \cdots)\ dx^1\wedge dx^2\wedge\cdots\wedge dx^n,
$$  
where $\exp^\ast_p\bigl(\mathrm{Ric}(\nabla)\bigr)_p = R_{ij}\, dx^idx^j$.  Thus, Ric gives the deviation of the parallel volume form from the exponentially flat one.  (This makes sense, even though you can't define 'geodesic balls' without a metric.  You still compare the volume of open neighborhoods of $p$ with respect to the two 'natural' volume forms.)