How wide is the Birkhoff Polytope? Now also posted on Math Overflow.

Define the width of a polytope $P \subset \mathbb R^d$ as the minimum length of the interval $\{v \cdot p:p \in P\}$ for $v$ in the unit sphere. In other words the width is the smallest number $W$ such that you can sandwich $P$ between two hyperplanes distance $W$ apart. Here's a picture:
 
Suppose the polytope $P \subset \mathbb R^d$ is contained in the affine subspace $A + x$ for $A \subset \mathbb R^d$ a hyerplane. Define the relative width as the smallest length of $\{v \cdot p:p \in P\}$ as $v$ ranges over the unit sphere in $A$. In other words translate the affine subspace to contain the origin and then ignore the perpendicular directions.
The Birkhoff polytope $\mathcal B$ is defined as the convex hull of the $n!$ permutation matrices. That means the $n \times n$ matrices with all zeros except for exactly one $1$ in each row and column. Equivalently $\mathcal B$ is the set of nonnegative matrices with all row and column sums equal to $1$.
In this case the affine subspace is defined as
$$\left \{x \in \mathbb R^d: \sum_j x^i_j =1, \sum_i x^i_j =1\right \}.$$
This just says the row and column sums equal $1$. Within that subspace the polytope is defined as the intersection with the first quadrant.
I am having trouble computing or estimating the height of $\mathcal B$. I would imagine the $v$ that minimises the projection is something like
$$
   v =
  \left( {\begin{array}{cccc}
   1/4 & -1/4 & 1/4& -1/4\\
   -1/4 & 1/4 & -1/4 & 1/4\\
   1/4 & -1/4 & 1/4 & -1/4\\
  - 1/4 & 1/4 & - 1/4 & 1/4\\
  \end{array} } \right)
$$
or in general make half the diagonals equal to $1/n$ and the other equal to $-1/n$. Then choosing the correct permutation matrices for the endpoints of the interval, we can force the interval to have length $2$.
The only reason I have to believe this is there are many choices of permutation matrices, and we want to minimise the interval length among all pairs. So $v$ should be symmetric in some sense.
Does anyone have ideas?

Have you tried using Lagrange Multipliers?
Define a function $F: S_{n \times n} \to \mathbb R$ by $$F(x) = \max\{|x \cdot(a-b)|: a,b \text{ vertices of } \mathcal B\}$$
To show $v \in S_{n \times n}$ is a minimiser it is enough to show the subgradient at $v$ contains a vector normal to the sphere. Namely $v$ itself. The subgradient of $F= \max\{f_1,\ldots, f_N\}$ at the point $x$ is the convex hull of
$$\{\nabla f_i (x): f_i(x)=F(x)\}$$
Since $i$ runs over pairs of vertices it is straightforward to see $f_i(v)=F(v)$ iff $a$ and $b$ are in some positive and negative diagonal of $v$.  By symmetry add up all the $\nabla f_i (x)$ to get a positive multiple of $v$ and done!
