If $X$ is a simplicial complex, is there a characterization of the links of its vertices that is equivalent to the statement "$|X|$ is a manifold"?

I don't think there will be a simple necessary and sufficient condition.

For exotic triangulations, the links $L$ of vertices will not be manifolds. One thing you need (not sufficient) is for $L \times \mathbb{R}$ to be a manifold. It's kind of shocking that this is possible for non-manifolds; the first example of it is Bing's dogbone space. A weaker version of the phenomena is the Whitehead manifold $W$, which is a contractible $3$-manifold which is not homeomorphism to $\mathbb{R}^3$ but such that $W \times \mathbb{R}$ is homeomorphism to $\mathbb{R}^4$.

To get an idea of the kinds of things that are involved here, I recommend reading the introduction to Edwards's paper "Suspensions of homology spheres", available here. That paper also contains a very detailed account of the fact that the double suspension of Mazur's homology 3-sphere is a 5-sphere (with lots of pictures). Another nice account of this is in Steve Ferry's notes on geometric topology, available here.

In dimensions $\geq 5$, Theorem 1.5 of Galewski and Stern implies that if the links of the vertices of a homology manifold of dimension $\geq 5$ are simply-connected, then it is a manifold.

I'm not sure if this is equivalent to your condition. Certainly your condition implies that the links of vertices are simply-connected and are homology spheres. But to be a homology $n$-manifold, the links of the barycentric subdivision must be $n-1$-homology spheres as well, and I'm not sure whether that is implied by your conditions.

In $1, 2, 3$ and (I believe) $4$ dimensions, I think it's necessary and sufficient to have links be $n-1$-spheres.