It is necessary and sufficient that for any nonzero $d\in\Bbb Z/n\Bbb Z$, there exists $i\in\{1,2,\dots,k\}$ such that $d\notin (A_i-A_i)$. In other words,
$$\bigcap_{i=1}^k (A_i-A_i) = \{0\}.$$
This holds even for sets of varying sizes.

-----

Since $| A_i-A_i|\geq |A_i| = m$, we get a necessary condition: $n-1\leq k(n-m)$, that is
$$m\leq \frac{(k-1)n+1}k.$$
For varying set sizes, it is
$$n\geq \frac{m_1+\cdots+m_k-1}{k-1}.$$

----

Another necessary condition can be obtained from the observation that for any $a\in\Bbb Z/n\Bbb Z$, there exist $m^k$ vectors $(c_1,\dots,c_k)$ such that $a\in \bigcap (A_i+c_i)$. Since these vectors must be distinct for distinct $a$, we have $n\cdot m^k\leq n^k$, that is
$$m\leq n^{(k-1)/k}.$$
This condition implies that the given example for $k=2$ is optimal when $m^2\leq n<(m+1)^2$.

For varying set sizes, the last condition takes form:
$$n\geq (m_1\cdots m_k)^{1/(k-1)}.$$