Is there some analogy for the phenomenon of chirality related to enantiomorph polyhedrons, but where the family of enantiomorphs is more than two polyhedra, in three or more dimensions, and related through C3 symmetry instead of a mirror image ?
C3 symmetry, chirality and polytopes
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There is a gradual deformation of an icosahedron where the enantiomorph deformation on faces is C3 (and C5 on vertices and C2 on edges):
It is used to explain $Au_{60}$. Could the bond distribution in a chiral $C_{60}$ buckminsterfullerene be similar? Eventually but there are also chiral truncated pseudo tetrahedrons where the truncation area lacks mirror planes which might explain point symmetrical chirality. Such an arrangement can also be made from making four similar chiral selections of faces in an icosahedron. In flatted form:

"more than two polyhedra"? A polyhedral collection can of course be cut or joined further, like from the image above. I think the question would benefit by specifying cases that don't qualify as answers, if you mean so.
Not really. The existence of chirality comes abstractly from the fact that the orthogonal group $O(n)$ has exactly two connected components separated by the sign of the determinant, namely the rotations (determinant $1$) and the reflections (determinant $-1$). This means that if you take a shape in $\mathbb{R}^n$ and apply every element of $O(n)$ to it you generically get two disconnected families of shapes, each of which is the reflection of the other. If two shapes are related by a rotation then you can just move one of them until it exactly overlaps the other but you generally can't do that with a reflection.
So the $C_2$ is not something we impose, it's already there. Even if you want to consider more general linear transformations, $GL_n(\mathbb{R})$ also has exactly two connected components which are also separated by the sign of the determinant ("orientation-preserving" vs. "orientation-reversing"). I think the same phenomenon persists even if we go all the way up to the homeomorphism group $\text{Homeo}(\mathbb{R}^n)$ but I'm not sure.