Why do convex polytope options constrict with dimension, rather than expand? There are an infinite number of regular polygons in the plane,
five regular polyhedra,
six regular polytopes in $\mathbb{R}^4$,
and then three regular polytopes in every dimension $d > 4$.
There are eight convex deltahedra (all faces equilateral triangles),
five in $\mathbb{R}^4$ (see, "Convex deltahedra in higher dimensions"),
and then three deltatopes in every dimension $d > 4$.

Q. Is there some intuitive reason why freedom is removed in higher dimensions?

I ask because, if I didn't know better, I would think the opposite.
There is vastly much more "room" in higher dimensions, and one might think forms proliferate, even under constraints. Concerning "much more room,"
think of the severe contraints on planar graphs vs.
the fact that every graph can be realized as embedded in $\mathbb{R}^3$.
I seek a corrective to my faulty intuition. Thanks!
 A: This is a well known "dimension curse" phenomenon. It is easier to explain for spherical polyhedra.  Let $P$ be a regular spherical convex polyhedron of dimension $d$. Starting with dimension $d=3$, all spherical convex polyhedra are rigid, so they are determined by their links (intersection of a vertex cones with an $\epsilon$-sphere).  The latter are themselves regular (spherical) polyhedra.  Therefore, for $d\ge 4$ these links are rigid themselves and regular polyhedra are uniquely determined by their links (which are all congruent, of course, by regularity).  Thus, as dimension grows the combinatorial types of regular polyhedra can disappear but new cannot appear.  
For Euclidean polyhedra the same type of analysis works, but goes through the links which are spherical polyhedra.  Therefore, as regular spherical polyhedra disappear so do Euclidean ones, but this is less intuitive perhaps.  I haven't seen "deltatopes" before, but it's the same story there as well, if I understood definitions correctly. 
