I found this on C|NET, and wondering if there was anything like a $1000\times1000\times1000$ cube? And how many arrangements would it have?
Or if this is too much, how about an asymptotic formula for $R(n)=$ number of permutations of an $n\times n\times n$ cube?
Cubes with odd and even side lengths behave slightly differently, because an odd cube have side-center cubies that remain fixed in space. But let's consider an even cube, since 1000 is even. It will be convenient to write the side length as $1000=2+2k$.
Let's first get an overview of the orbits of the small visible cubies.
Corner pieces form one orbit of 8 cubies. On an even cube those are the only ones that can be twisted in place -- an edge or side piece always "knows" mechanically what the direction towards the center of the edge, respectively the side, is. Just as in the ordinary 3×3×3 cube, the number of corner orientation states is $3^7$.
Edge pieces form $k$ orbits of 24 cubies each (each orbits contains two positions on each edge, grouped by their distance from the middle).
Center pieces form $k^2$ orbits of 24 cubies each, namely one for each position in a quadrant of each side. Each orbit contains four positions on each side. $k$ of these orbits are special because they are on the diagonal on the side, and so their pieces can "see" only one kind of edge pieces.
For each of these $1+k+k^2$ orbits, we can see, given a bit of experience with a 4×4 cube, that at least every even permutation can be realized without upsetting the other orbits. We'll need to figure out which (or at least how many) combinations of odd permutations are possible.
A quarter twist of a side layer is an odd permutation of the corner orbit, and of each side orbit, but is an even permutation each edge orbit (namely two 4-cycles)
A quarter turn of an inner slice is an odd permutation of the single edge orbit it touches, and an odd permutation of each side orbit it touches, except that the diagonal side orbit it touches gets an even permutation (namely, again, two 4-cycles). It doesn't touch any corners, of course.
These two cases can be combined to form any legal move, so this means that if we know the parity of all of the corner and edge orbits (which can be chosen freely), the parity of each side orbit is uniquely determined.
Thus, we now have the following factors: $$ \frac{3^7 \cdot 8! \cdot (24!)^k \cdot (24!)^{k^2}}{2^{k^2}} $$ where the denominator represents the fact that the parity of the center permutations is fixed.
We need to take care of some overcounting, though.
First, there are legal moves that turn the entire cube as a unit, slice by slice, but the result of that probably shouldn't count as a different position. So we need to divide by the number of orientation-preserving symmetries of a cube, which is 24.
Second, some of the side pieces are indistinguishable. More precisely, each orbit of 24 side pieces splits into 6 groups of 4 indistinguishable pieces. We can permute each group of 4 freely without the result being visible, except that the total parity of what we do to each color must be even (because the parity of the entire orbit is fixed by what the edges and corners look like). That gives for each orbit a factor of $(4!)^6/2$.
(The edge pieces are all distinguishable -- for each pair of neighboring colors, each orbit contains two edge pieces with these colors, but they are mirror images of each other and can therefore not be swapped unseen).
The final count of visually distinguishable positions on a $(2+2k)^3$ cube is therefore $$ \frac{3^7 \cdot 8! \cdot (24!)^k \cdot (24!)^{k^2}}{2^{k^2} \cdot 24 \cdot ((4!)^6/2)^{k^2} } = \frac{3^7 \cdot 8!}{24} (24!)^k \biggl(\frac{24!}{(4!)^6}\biggr)^{k^2} $$ Note that the initial factor is the number of positions of a 2×2×2.
For a $(3+2k)^3$ cube a similar analysis gives $$ \frac{3^7 \cdot 8! \cdot 2^{11} \cdot 12!}{2} (24!)^k \biggl(\frac{24!}{(4!)^6}\biggr)^{k(k+1)} $$ where the initial factor is the number of positions of a 3×3×3.