For any given order $n$, find the length of the longest contiguous subsequence that occurs in every distinct binary de Bruijn sequence of that order. We'll call that $L(n)$.
We define distinct de Bruijn sequences as those that differ regardless of reflection and rotation. For example, 00011101 and 00101110 would be considered the same sequence; thus for $n=3$, our answer would be $L(3)=8$, the length of that single distinct sequence. Likewise, we have the trivial cases $L(1)=2$ and $L(2)=4$, but that pattern won't keep up.
For larger $n$, we can still make progress. It's trivially true that $L(n) \geq n$, since every sequence of length $n$ must appear in $B(2,n)$ by definition.
We can slightly improve that naïve lower bound by observing that every sequence of order $n$ must contain $n$ ones in a row, padded by at least one zero on either side (since $n+1$ ones can never appear in an order-$n$ sequence). Therefore there exists a sequence of at least length $n+2$ which is common to every de Bruijn sequence of order $n$, so $L(n) \geq n+2$ for all $n>1$.
My guess is that that bound can be improved; I wouldn't be surprised if there's a simple closed-form expression for $L(n)$. If you can either find such a function, or alternatively give some reasoning as to why this question actually turns out to be difficult, I'll consider this answered.