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Take $B$ to have $1$s on the main diagaonal and the first superdiagonal, then $\|B\|\leq 2$. If we take $C$ to have $1$s on the diagonal and filling the last column, then $\|C\|\geq \sqrt{n}$. (These both have full rank and $2n+1$ $1$'s.) This shows that $\alpha_n\geq \frac{\sqrt{n}}{2}$, though from the examples in Robert Israel's answer this bound is not optimal.