I found a neat way to get from the permutohedron of order n to the associahedron of order n: Associate an ordered (height-wise) tree to each vertex, edge, face, etc... to the permutohedron of order n. Then associate an unordered tree to each vertex, edge, face, etc... to the associahedron of order n. By forgetting the ordering, we get a natural map from the permutohedron of order n to the associahedron of order n. I demonstrated how this works in the case n=4, in the image above: Simply collapse each red edge to get from the permutohedron of order 4 to the associahedron of order 4. There seems to be a lot of interesting combinatorics here. What I have found so far is that the symmetric group acts as reflections on the permutohedron of order n via its action on the ordered trees, and elementary matrices act on the associahedron of order n via its action on the unordered trees.
In the image above, you can see four sets of three vertices on the permutohedron of order 4 collapsing to a single vertex on the associahedron of order 4, two sets of two vertices on the permutohedron of order 4 collapsing to a single vertex on the associahedron of order 4, and eight sets of one vertex on the permutohedron of order 4 collapsing to a single vertex on the associahedron of order 4. So we get a partition of 4! as
$4! = 8 \cdot 1 + 2 \cdot 2 + 4 \cdot 3 $
I want to know how this generalizes for various n. So how do the vertices of the associahedron of order n partition the vertices of the permutohedron of order n? The answer seems quite involved. For example, there are
$ 2^{26} \cdot 3^4 \cdot 5^5 \cdot 11^2 \cdot 13^2 \cdot 17 \cdot 19 \cdot 23 \cdot 29 $
vertices on the permutohedron of order 15 which collapse to some vertex on the associahedron of order 15.