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.

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.

EDIT: For example when $n=5$:

\begin{equation*} B= \begin{pmatrix} 1 & 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 1& 1 \\ 0 & 0 & 0 & 0 & 1\end{pmatrix} , \quad C= \begin{pmatrix} 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1\end{pmatrix} \end{equation*}