Answer: 592

Small Hint:

Divisibility by $5$ determines the last digit

Big Hint:

Divisibility by $3$ restricts the number of digits equal to $8$

Solution:

A multiple of $15$ must end in $0$ and have digit sum divisible by $3.$ Because $8\equiv2\pmod3,$ the number of $8$’s must be a positive multiple of $3.$ The smallest possible number therefore has three $8$’s followed by $0,$ namely $n=8880.$ Thus $\frac{n}{15}=592.$