There's a neglected part of this puzzle: Proving there's a path for the inner hypercube, through the outer hypercube. Confirming that the left over piece of the OH is connected, and that there's an actual tunnel through it. Perhaps showing that the IH can pass through the tunnel -- if it's not straight, maybe there are twists and turns?
We can check the 3D case with a model. But the higher dimensional margins are pretty skimpy, suggesting it might not be a cakewalk.
Rich
--if there ever were a problem that could be described as a "cakewalk" this must be it! But yes, based on the literature Gosper mentioned (I had no idea but this cubes-thru-cubes problem goes back to 1693!) it seemed like nobody had considered the possibility of a twisting/turning tunnel? It seems "obvious" that best is a straight tunnel, but it seemed like nobody had proved that, and thus perhaps by twisting/turning one could optimize the cube size ratio a little more than one could with a straight tunnel? No... I think not. See http://mathworld.wolfram.com/CubeSquareInscribing.html especially picture, which seems to show, that the size ratio is determined by a single moment in the history of the cube-passing-thru-tunnel. Therefore, the twisting and turningness of the tunnel is irrelevant to improving the size-ratio. Also the connectedness of the cube after drilling the tunnel, is "obvious" in all dimensions>=3. If so, the problem is equivalent to the question of what is the largest (n-1)-hypercube you can inscribe in an n-hypercube. If it is believed (which happens to be true when n=1,2,3) that the answer involves the two hypercubes having the same center, then the answer arises from an nXn rotation matrix, times a scaling factor. This has n*(n-1)/2+1 degrees of freedom. [The other n*(n+1)/2 degrees of freedom are not free, they are determined by simultaneous quadratic equations saying the rows of matrix are orthonormal.] You optimize those degrees of freedom subject to the demands that all vertices of the rotated cube, lie inside or on the boundary of the unrotated cube, which is a set of linear inequalities. (If we did not believe the "same centers" hypothesis, then an additional n degrees of freedom would be needed.) So, whatever the optimum is, is determined by n*(n+1)/2 quadratic equations plus some linear equations, involving n^2+1 real variables. We can eliminate all variables but 1 by the method of "Macaulay multipolynomial resultants" which appears to be not very well understood even over 100 years after F.S.Macaulay invented them. He wrote a book "The algebraic theory of modular systems" (Cambridge University Press, 1916) available online here https://archive.org/details/algebraictheoryo00macauoft More pedestrianly, we can eliminate one variable at a time by computing pairwise resultants with respect to that variable, for pairs of polynomials, using one of several standard determinant formulas (one is by Sylvester) for the polynomial-pair resultant. That means if we start with P polynomials (in P variables) of degree <=d then after we eliminate a variable we get (P-1)*P/2 polynomials (in P-1 variables) of degree <=d*d (I think; perhaps a somewhat greater degree bound is necessary) and I presume we may discard all but P-1 of those polynomials then continue on. This repeated degree-squaring explains the tremendous degree growth in the algebraic numbers. After eliminating all but 1 variable, P-1 eliminated in all, the degree is squared P-1 times, resulting in final degree<=d^(2^(P-1)) for the final univariate polynomial. In our cubes-thru-cubes problem in n dimensions this would predict the algebraic number answer would be of algebraic degree about 2^(2^(n*(n-1)/2)), note the DOUBLE exponential. Now by the use of Macaulay rather than pedestrian resultant techniques, and being actually careful (I've been non-careful), one can probably improve the degree bounds versus what I just said. Further, there could be miracles where, e.g, the polynomials magically factor, or decompose ala Kozen-Landau, so you have less degree than naively thought. But it should be clear the algebraic degrees probably are going to be pretty damn enormous pretty damn soon, and I very much doubt enough miracles can occur to save you from that. Hence I also think, above some moderate dimension (say 10), it is going to be utterly hopeless to express the optimum answer to the cube-thru-cube problem in any kind of closed form that fits in the observable universe. Which returns me to the question I posed as "more interesting" and certainly more hopeful: what is the asymptotic behavior of the cube-thru-cube answer in large dimension n? And wikipedia claims one can pass a regular simplex thru a hole in another, and a regular octahedron -- both those problems again can be generalized to n dimensions. John Wallis in the first ever solution (which was suboptimal) in 1693 of the cube-thru-cube problem, assumed the direction of motion was the long diagonal. Because a cube viewed that way projects to a regular hexagon, Wallis found largest square inscribable in reghex, and voila. Can we use Wallis's idea to derive a lower bound for the n-dimensional problem? Well. It seems to me if you were to drill a conventional circular (cross section is hyperspherical) hole thru the hypercube [-1/2,1/2]^n in the (1,1,1,...,1) direction, then the largest radius R you could use would obey: R is the length of the vector from 0 to the mean of (n-1) hypercube vertices. So I think 2R = squareroot( n - log2(n) ) approximately. That is not quite large enough, we would need sqrt(n-1) for the most naive lower bound from the circular hole idea, to get traction. So, the most naive Wallis-like attack failed.