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, consider the first Bianchi identity: $R^i_{jkl}+R^i_{klj}+R^i_{ljk}=0$.  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.

Assume now 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 $\exp^\ast_p\bigl(\mathrm{Ric}(\nabla)\bigr)_p = R_{ij}\, dx^idx^j$.  (Here, the $x^i$ are any linear coordinates on $T_pM$ centered at $0_p$ that are $\Upsilon$-unimodular at $0_p$.)  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.)