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.

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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
$$n\geq \frac{m_1+\cdots+m_k-1}{k-1}.$$

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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:
$$n\geq (m_1\cdots m_k)^{1/(k-1)}.$$

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As for the construction, the following streamlining of Gerhard's idea does the job for a given $k$: 

Take any integers $0<b_1<b_2<\dots<b_k$, set $n:=\mathrm{LCM}(b_1,\dots,b_k)$, $m:=\frac{n}{b_k}$, and select $A_i$ as any subset of $b_i(\Bbb Z/n\Bbb Z)$ with $|A_i|=m$. Indeed, in this setting, for any nonzero $d\in\Bbb Z/n\Bbb Z$, there exists $i\in\{1,2,\dots,k\}$ such that $b_i\nmid d$, implying that $d\notin (A_i-A_i)$.
For Gerhard's example with $k=3$ and $b_1=5$, $b_2=6$, and $b_3=7$, we have $n=210$ and $m=30$.

If we are also given $n$, this construction becomes more tricky as we need to pick $k$ its divisors $b_1<b_2<\dots<b_k$ with the smallest $b_k$ possible such that $n\mid \mathrm{LCM}(b_1,\dots,b_k)$. It is clear that $b_k$ cannot be smaller than the largest primepower dividing $n$. It also follows that here it does not make sense to have $k$ greater than the number of distinct primes dividing $n$, since $b_i$ being the distinct primepowers forming the prime factorization of $n$ do the job.