If one considers the distinguishing feature of the sign homomorphism $S_n \to \mathbb{Z}/2$ to be that it is the canonical map from $S_n$ to its abelianization, then there is nothing analogous for $S_\infty$ in the sense that the abelianization of $S_\infty$ is trivial. The abelianization of a group $G$ is also the group homology $H_1(G, \mathbb{Z})$, and in fact for $G = S_\infty$, all the homology groups $H_i(S_\infty, \mathbb{Z})$ vanish for $i > 0$; $S_\infty$ is an acyclic group. See Acyclic groups of automorphisms whose first page contains a statement of this result and similar ones.

If one considers the distinguishing feature of the sign homomorphism $S_n \to \mathbb{Z}/2$ to be that it is the canonical map from $S_n$ to its abelianization, then there is nothing analogous for $S_\infty$ in the sense that the abelianization of $S_\infty$ is trivial. The abelianization of a group $G$ is also the group homology $H_1(G, \mathbb{Z})$, and in fact for $G = S_\infty$, all the homology groups $H_i(S_\infty, \mathbb{Z})$ vanish for $i > 0$; $S_\infty$ is an acyclic group. See Acyclic groups of automorphisms whose first page contains a statement of this result and similar ones.