The wikipedia's page for Birkhoff polytope states that the polytope has $n^2$ facets, determined by the inequalities $x_{ij} \geq 0$, for $1 \leq i,j \leq n$. I've tried different things but can't see how this claim makes sense. How should I look at this?
Also is this only true for $n > 2$? According to the Birkhoff - von Neumann theorem, $B(2)$ would have dimension $(n-1)^2 = 1$, $n! = 2$ vertices, but also $n^2 = 4$ facets, which are also vertices?
The penultimate sentence of the Wikipedia article is referring to half-spaces of the $(n-1)^2$ dimensional space containing $B_n$. $B_n$ is the intersection of these half-spaces. The formula for the facets themselves is $a_{i,j} = 0$ for each of the $n^2$ entries.
And, yes, the number of facets of $B_n$ is $n^2$ only for $n \gt 2$. The Wikipedia article is not stated correctly. As you have deduced, $B_2$ is a line segment.