Smallest regular simplex containing the unit cube in $R^n$

What is the length $e_n$ of the edge of the smallest $n$-dimensional regular simplex $S_n$ containing the $n$-dimensional unit cube $Q_n$?

In particular, is there $n$ such that $e_n<\sqrt{2}(n+1-\sqrt{n+1})$?

The question is motivated by the question Two cubes in unit cube. A positive answer to the second part would (almost) imply that two cubes with side of length greater than $1/2$ would fit inside the unit cube without overlap. Where "almost" means "if the centers of the cube and the simplex are not too far away": the plan would be to cut the unit cube by a hyperplane orthogonal to a main diagonal and fit the small cube inside each of the two halves.

Since $Q_n$ contains a ball $B_n$ of diameter $1$, the simplex $S_n$ also contains $B_n$. Therefore, the altitude of $S_n$ is at least $(n+1)/2$ and so $e_n \ge \frac{1}{\sqrt 2} \sqrt{n(n+1)}$.

On the other hand, a "greedy" recursive construction (where a facet of $Q_n$ is contained in a facet of $S_n$) gives the upper bound $e_n \le \sqrt{2}\left(\sqrt{\frac{1}{2}} + \sqrt{\frac{2}{3}} + \cdots +\sqrt{\frac{n}{n+1}}\right) = \sqrt{2}n - \Theta(\log n)$.

The three-dimensional case hase been completely solved: Ogilvy and Robbins, 1976, Croft, 1980

It has been shown that the greedy construction is indeed optimal in $\mathbb{R}^3$; however, two other locally minimal regular tetrahedra are larger just by a few percent.

This paper by M. Nevskii is relevant. It is shown there that a simplex contains a translate of $[0,1]^n$ iff $\sum_i\frac1{d_i}\leq 1$, where $d_i$ is the length of the longest segment inside the simplex which is parallel to the $i$th axis. In particular, this clearly implies $e_n\geq n$ (and this is far from being sharp).