The Banach-Mazur distance $d(X, Y)$ between two normed spaces $X, Y$ of the same dimension is defined as $d(X, Y) = \log\inf \|T\| \cdot \|T^{-1}\|$, where the $T:X \to Y$ is a linear and invertible operator. The estimates between classical $\ell\_p^n$ spaces are known and in particular we have that $d\left(\ell\_1^n, \ell\_{\infty}^n\right) \sim \sqrt{n}$ asymptotically. Using Walsh matrices it can be shown that for $n=2^k$ this distance is actually not greater than $\sqrt{n}$.


In the two dimensional case we clearly have $d(\ell\_1^2, \ell\_{\infty}^2)=1$. I wonder if anybody has managed to provide some more specific results in the three dimensional case. It is equivalent to finding two positively homothetic parallelotopes with one contained in and the second containing the regular octahedron with the homothety ratio as minimal as possible. It seems like pretty basic question in three dimensional discrete geometry, but still I could not find any information and the problem might be highly non-trivial. The only information I found was in the paper of Stromquist ("The maximum distance between two dimensional Banach spaces") where he wrote that this distance is not known. But it was 35 years ago, so I guess that there could be some new results in this direction. If not, maybe at least one could make some good conjecture on what this distance should be like?



Banach-Mazur distance between the cube and the octahedron
This can be done as a nonlinear optimization problem: $T$ is a $3 \times 3$ matrix, and if $e\_i$ and $v\_j$ are the vertices of the unit balls in $\|\cdot\|\_1$ and $\|\cdot\|\_\infty$ norms respectively, you want to minimize $s$ subject to constraints $\|T e\_i\|\_\infty \le 1$, $\|T^{-1} v\_j\|\_1 \le s$ (or equivalently, $\|\text{Adj}(T) v\_j\|\_1 \le s |\det(T)|$, with $\det(T) \ne 0$).  

Maple's Global Optimization Toolbox returns (almost immediately) an approximate solution, which is (up to roundoff error)


$$ T = \pmatrix{1/3 & 1 & 1\cr -1 & 1 & -1/3\cr 1 & 1/3 & -1\cr},\ s = \dfrac{9}{5}$$


It is easy to check that this does satisfy the constraints.
Thus it appears that the answer is $9/5$. It should be possible to prove optimality rigorously, if somewhat tediously, using the Karush-Kuhn-Tucker conditions.


EDIT: I also tried the $4$-dimensional case. The solution Maple obtained was


$$ T = \pmatrix{0 & 1 & -1 & 1\cr -1 & 0 & 1 & 1\cr 1 & -1 & 0 & 1\cr -1 & -1 & -1 & 0\cr}, \ s = 2 $$


