Answer: The smallest value is $p+2$ for $p$ odd and $2$ for $p=2$.

The $p=2$ case is easy so we assume $p$ is odd.
Take $A$ to be the set $\{1, 2, \dots, p-1, p+1, 2p-1\}$. Then $A$ is a $p$-set and the sum of all elements of $A$ is divisible by $p$. Hence the sum of any $p$ elements in $A$ is not divisible by $p$. Thus $n \ge p+2$.

Now let $A$ be a $p$-set with at least $p+2$ elements. We may assume $A = \{a_1, \dots, a_{p-1}, a, b, c\}$, where $a_i \ne a_j \pmod p$ for $1 \le i \ne j \le p-1$. Let $S := \sum_{i=1}^{p-1} a_i$ and let $T := a+b+c+S$.

If $T = a + a_i \pmod p$ for some $i$, then the remaining $p$ elements have sum divisible by $p$ and we are done. Thus we may assume that $T \ne a + a_i \pmod p$ for all $i$. Then we have
$$T + \sum_{i=1}^{p-1} (a + a_i) \equiv 1 + 2 + \dots + p \equiv 0 \pmod{p}.$$
It follows that $a \equiv S+T \pmod p$. Similarly, we may assume $b \equiv c \equiv S+T \pmod p$. Then
$$2 \left( a + \sum_{i=1}^{p-1} a_i \right) \equiv 2a + 2S \equiv a + b + c + S - T \equiv 0 \pmod{p}.$$

This gives a subset with $p$ elements with sum divisible by $p$.