A compact orbifold without boundary will have a fundamental chain in rational singular homology if and only if it is orientable.  The fundamental chain will satisfy the same sorts of properties as for a manifold.  In particular, collapsing the complement of a small disc will send the fundamental chain to the generator of $H_n(D^n,\partial D^n)$.  The reason that closed oriented orbifolds have fundamental chains in rational homology is essentially because, to the eyes of rational homology, they look like manifolds.

So the first part of what you are asking amounts to the question of whether $\overline{\mathcal{M}}_{g,n}/S_n$ is orientable or not.  

Just so we're clear, $\overline{\mathcal{M}}_{g,n}$' is the moduli space of stable curves of genus $g$ with $n$ marked points labelled $1 \ldots n$.  The action of the symmetric group $S_n$ is by permuting the labels of the marked points.  Before taking the quotient by $S_n$, the space $\overline{\mathcal{M}}_{g,n}$' is a complex orbifold, so the complex structure induces a well-defined orientation just as for complex manifolds.

So the question is now whether the $S_n$ action preserves the orientation or not.  The answer is that it does indeed.  The symmetric group action is in fact algebraic, so it preserves the complex structure and hence the orientation.

As an aside: There is a homotopy equivalence between the moduli space $M^{rib}_{g,n}$ of metric ribbon graphs (that thicken to a genus $g$ surface with $n$ labelled boundary components) and the uncompactified moduli space $\mathcal{M}_{g,n}$.  The action of $S_n$ on $M^{rib}_{g,n}$ is **not** orientation preserving.  This is because, via Strebel differentials (or Penner's Lambda lengths) one can show that $M^{rib}_{g,n}$ is homeomorphic to $\mathcal{M}_{g,n} \times \mathbb{R}_+^n$ and under this homeomorphism the symmetric group action corresponds to the usual action on the first factor and the permutation action on the coordinates of the second factor.  Thus it doesn't act in an orientation preserving way.