If the third powers count the packing of a cube: [1, 8, 27, 64, 125, 216, 343..]
And the first differences of the fourth powers form a cubic sequence that counts a packing of a rhombic dodecahedron (see picture): [15, 65, 175, 369, 671, 1105, 1695..].
Do the second differences of the fifth powers have any associated polyhedra with a packing counted as: [30, 180, 570, 1320, 2550, 4380, 6930..]?
What about the cubic differences of the nth powers?
That sequence doesn't count the volume of the rhombic dodecahedron. The volume is given by $\frac{16\sqrt{3}}{9}a^3$. But from the image you linked it does seem to count some kind of sphere packing. The $n^4-(n-1)^4$ seems to come from the fact that some vertices have $4$ edges connected to it which can be used to generate the whole polyhedron with the parametric equation:
$$ \vec{p}(t_1,t_2,t_3,t_4) = t_1\vec{e_1} + t_2\vec{e_2} + t_3\vec{e_3} + t_4\vec{e_4} $$
where the $\vec{e_k}$'s are the vectors corresponding to the edges and $0\le t_k \le 1$. Each edge multiplies the total number of spheres by $n$, except that the $4$th time you do this, you're counting twice a total of $(n-1)^4$ spheres.
Another way to think of this is by using the inclusion–exclusion principle. We start by splitting the rhombic dodecahedron into $4$ triagonal trapezohedra, each having $n^3$ spheres, ignoring the overlap on the faces. To take this overlap into account, we must remove $6\times 4-12 = 12$ faces, with a "total" of $12 n^2$ spheres, but now we counted the spheres on some edges twice. There are $4$ such edges that we must add back, but if we add $4n$ spheres now it's the center sphere that was counted twice and we must remove. Adding it all up gives $4 n^3 - 6 n^2 + 4 n - 1 = n^4-(n-1)^4$.
Can we do something similar for another polyhedron and get the second differences of the fifth powers? I suspect not, otherwise I think this result would be noteworthy enough to show up on some quick searches (specially on the OEIS page). The most obvious generalization would come with the rhombic triacontahedron, because it can be split into $20$ trigonal trapezohedra and it can be generated by a similar parametric equation:
$$ \vec{p}(t_1,t_2,t_3,t_4,t_5) = t_1\vec{e_1} + t_2\vec{e_2} + t_3\vec{e_3} + t_4\vec{e_4} + t_5\vec{e_5} $$
The second difference of the fifth power is $20 n^3 - 60 n^2 + 70 n - 30$, which looks promising because of the $20n^3$, but the next term should have been $-(6\times 20-30)n^2 = -90n^2$ to correspond with the $90$ faces that were counted twice. So this doesn't seem to work. The first method (using the parametric equation) also doesn't work because the overlap begins with $n^4$, and not $n^5$.