Solution:
$$\frac{5}{7}$$
Notice that the answer is the sum of the probabilities that boxes $1,2, \ldots$, respectively, contain at least 2 balls, since those events are mutually exclusive. For box $n$, the probability of having at least 2 balls is
$$3\left[\left(1 / 2^{n}\right)^{2}\left(1-1 / 2^{n}\right)\right]+\left(1 / 2^{n}\right)^{3}=\frac{3}{2^{2 n}}-\frac{2}{2^{3 n}}=\frac{3}{4^{n}}-\frac{2}{8^{n}}$$
Summing to infinity using the geometric series formula, we get the answer $\frac{3/4}{1-1/4}-\frac{2/8}{1-1/8}$, which is equal to $\frac{5}{7}$.