Every dual $T^*$, where $T$ is torsion-free -- and hence every reflexive module -- is a second syzygy, as displayed by dualizing a projective presentation of $T$. On the other hand, it follows from Auslander-Bridger (or see a slightly more readable presentation by Masek (last Corollary in this paper) that if a ring $R$ satisfies S1 and is Gorenstein at the minimal primes, then every second syzygy is reflexive.
Therefore, for reduced rings being a dual of a torsion-free is equivalent to being reflexive is equivalent to being a second syzygy. In particular, as long as the global dimension is at least 3, there are duals that are not projective.
Back to the original question: when can you conclude the dual of $T$ is free? (I'm going to talk only about local rings, so ignore the distinction between free and projective.) Assume $A$ is a regular local ring. If there is a module $N$ such that $\operatorname{Ext}(T,N)=0$ for $i = 1, ..., \operatorname{depth} (N)-2$, then the dual of $T$ is free. In particular, if $N$ has depth less than or equal to $3$ and $\operatorname{Ext}_R^1(T,N)=0$, then the dual of $T$ is free. This is in a recent paper by Jothilingam, but is not hard to prove directly. It's not a condition solely on $A$, but maybe it's useful.
