Let $C$ be the set of $n$ clowns. Label the colours $1,2,3, \ldots, 12$. For each $i=1,2, \ldots, 12$, let $E_{i}$ denote the set of clowns who use colour $i$. For each subset $S$ of $\{1,2, \ldots, 12\}$, let $E_{S}$ be the set of clowns who use exactly those colours in $S$. Since $S \neq S^{\prime}$ implies $E_{S} \cap E_{S^{\prime}}=\emptyset$, we have
$$\sum_{S}\left|E_{S}\right|=|C|=n,$$
where $S$ runs over all subsets of $\{1,2, \ldots, 12\}$. Now for each $i$,
$$E_{S} \subseteq E_{i} \quad \text{ if and only if } \quad i \in S,$$
and hence
$$\left|E_{i}\right|=\sum_{i \in S}\left|E_{S}\right| .$$
By assumption, we know that $\left|E_{i}\right| \leq 20$ and that if $E_{S} \neq \emptyset$, then $|S| \geq 5$. From this we obtain
$$20 \times 12 \geq \sum_{i=1}^{12}\left|E_{i}\right|=\sum_{i=1}^{12}\left(\sum_{i \in S}\left|E_{S}\right|\right) \geq 5 \sum_{S}\left|E_{S}\right|=5 n .$$
Therefore $n \leq 48$.

Now, define a sequence $\{c_{i}\}_{i=1}^{52}$ of colours in the following way:
$$\begin{array}{llllllllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ 4 & 1 & 2 & 3 & 8 & 5 & 6 & 7 & 12 & 9 & 10 & 11 \\ 3 & 4 & 1 & 2 & 7 & 8 & 5 & 6 & 11 & 12 & 9 & 10 \\ 2 & 3 & 4 & 1 & 6 & 7 & 8 & 5 & 10 & 11 & 12 & 9 \\ \end{array}$$
The first row lists $c_{1}, \ldots, c_{12}$ in order, the second row lists $c_{13}, \ldots, c_{24}$ in order, the third row lists $c_{25}, \ldots, c_{36}$ in order, and finally the last row lists $c_{37}, \ldots, c_{48}$ in order. For each $j, 1 \leq j \leq 48$, assign colours $c_{j}, c_{j+1}, c_{j+2}, c_{j+3}, c_{j+4}$ to the $j$-th clown. It is easy to check that this assignment satisfies all conditions given above. So, 48 is the largest for $n$.

Remark: The fact that $n \leq 48$ can be obtained in a much simpler observation that
$$5 n \leq 12 \times 20=240 .$$