All but the tetrahedron.
As noted in a comment, ultimately referencing a 1972 paper, for tetrahedra this cannot be done. I haven't looked at the paper, but the proof may go as follows: Let the vertices of your base tetrahedron be the standard basis vectors in $\mathbb R^4$ times 4, so their center point is the all-ones vector. Construct the four matrices which reflect through the faces of your tetrahedron (while fixing the sum-to-4 hyperplane). Observe that some entries of these matrices equal $\frac23$. Apply a reduced word in the reflection group to the all-ones vector, and prove by induction that the first generator in the word is always indicated by which entry has the lowest power of 3 in the denominator, with a predictable pattern mod 3 in the numerators. Conclude that no nontrivial product of reflections takes the center point back to itself.
For the octahedron, attaching a pair along opposite faces allows you to continue in the pattern of carbon atoms in the diamond crystal structure, where each atom has tetrahedral bonds in directions all four of which are the negatives of any neighbor's bonds. It is thus possible to glue together 12 octahedra positioned like the carbons and C-C bonds of the "chair" conformation of cyclohexane. Since eight of the twenty faces of an icosahedron are inclined like the faces of an octahedron, the exact same arrangement is possible with icosahedra.
EDIT: In fact, you can do better. Given any polytope that has two pairs of opposite parallel facets such that reflected polytopes may be attached to all four facets simultaneously without overlapping, a thickened parallelogram may be constructed by attaching eight identical polytopes along facets. This works because a pair of parallel reflections amounts to a pure translation, making the same possible attachment directions available at both ends of the double reflection.