I claim the answer is $4$, and is attained iff $n=4p$, where $p>2$ is a prime or when $n=8$. Check that indeed when $n=4p$, then $A_n=1+p$ whereas $B_n=2+4+2p$, yielding $B_n-2A_n =4$. Let $n=2^k\cdot m$, where $m$ is odd and $k\ge 2$. Notice that $A_n=\textstyle\sum_{d\mid m}d$. On the other hand, $B_n = 2A_n+4A_n+\cdots+2^k A_n - n$. Assume first that $k=2$. Then, $B_n-2A_n = 4A_n - n =4(A_n-m)$. Since $A_n\ge m+1$ (as $n>4$), it follows $B_n-2A_n\ge 4$. Equality holds iff $A_n = m+1$, that is when $m$ itself is an (odd) prime. Next, assume $k=3$. Then, $B_n=2A_n+4A_n+8A_n-n$, yielding $B_n-2A_n = 4A_n + 8(A_n-m)$. As $A_n\ge 1$ and $A_n-m\ge 0$, we find $B_n-2A_n\ge 4$ with equality iff $k=3$ and $m=1$, that is $n=8$. Finally, let $k\ge 4$. Then, $B_n = 2A_n + 4A_n + 8A_n +2^k A_n - n$, yielding $B_n-2A_n\ge 4A_n+8A_n\ge 12$, yielding a strictly worse value.