It's not clear from the linked article that the hexagon is a permutahedron, so we don't restrict to that case for starters. For our construction of higher dimensional scutoids we need a couple of standard polytope operations:
Prism: For a $d$-polytope $P$ let $\mathrm{prism}(P)$ be the $d+1$-dimensional prism over $P$ given by $P \times [0,1]$.
Vertex Truncation: For any vertex $v$ of $P$ let $E(v)$ be the edges of $P$ incident to $v$ ordered by length. Let $H$ be any affine hyperplane such that $v$ is the only vertex of $P$ lying on one side of $H$. Then the truncation $\mathrm{trunc}_{v,H}(P)$ is the subpolytope of $P$ that lies on the side of $H$ not containing $v$.
A $d$-scutoid is any polytope $Q$ that can be written as $Q = \mathrm{trunc}_{v,H}(\mathrm{prism}(P))$ for some $d-1$-dimensional polytope $P$ (called the base polytope) and some vertex $v$ of $P$. So a $3$-scutoid is combinatorially equivalent to a scutoid (in the sense of the linked article in the OP) when the base polytope is a pentagon.
If the above definition of a $d$-scutoid is not refined enough for some purpose one can always restrict to various families of allowable base polytopes. One chain of families that gets us from arbitrary polytopes to associahedra is $$\text{arbitrary polytopes} \supset \text{fiber polytopes} \supset \text{associahedra}.$$
Edit: As @Mark S points out in a comment below, scutoids are not polytopes: for scutoids to pack correctly some of their faces must be curved. (For visualizations see page 9 of the original article in Nature or this video.) Since the $3$-scutoids with pentagonal base polytopes described above are polytopes, they are technically not scutoids, but rather combinatorially-equivalent polytopal approximations.
Edit: An earlier version of this answer incorrectly identified the base polytope of a (combinatorial) scutoid as a hexagon instead of a pentagon.