I think answer is 11. We have: in any box has at least 4 balls.

Indeed, any box must contain at least $4$ balls; if a box contains just $3$ balls, then one with a same number must appear in at least $\lceil 7/3\rceil = 3$ other boxes, impossible. So we need at least $32$ balls, thus $m\geq 11$. A model for $m=11$ could be \[1234, 1579, 1689, 258A, 267B, 358B, 367A, 49AB4.\] With only $7$ boxes, the minimum $m$ would have been $7$, with unique model provided by the Fano configuration, equivalent to a symmetrc BIBD $(7,3,1)$.