Let $m=80$ and $n=12$. We will solve the problem for general $m $ and $n $. Label the people $a_i\forall 1\leq i\leq m $ and let $A=\{a_i:1\leq i\leq m\} $ be the set of all people, $B=\{1,2,...,n\} $. Denote the power set of any set $X $ by $\mathcal {P}(X) $. Define a function $f:A\longrightarrow \mathcal {P}(B)$ such that $j\in f (a_i) $ if and only if $a_i $ was called on the $j $-th day. What we require to find the murderer is that for any two people $a_i,a_j$ there is a day when $a_i $ was called and $a_j $ was not, and there is a day when $a_j $ was called and $a_i $ was not $\iff f (a_i) $ is never a subset of $f (a_j) \forall i\neq j$. We define a partial order $<$ on the set $\mathcal {P}(B) $ such that $X <Y;X,Y\in\mathcal {P}(B) $ if and only if $X $ is a subset of $Y $, then it follows that $\{f (a_i)\}_{1\leq i\leq m} $ is an anti-chain under the order $<$. By $\text {Sperner's Theorem} $, the length of the greatest anti-chain is $\binom {n}{\lfloor {\frac {n}{2}\rfloor}} $, hence it follows that the murderer can be found if and only if $m\leq \binom {n}{\lfloor {\frac {n}{2}\rfloor}} $. Back to our original problem, the above argument shows that even $9$ days suffice as $\binom {9}{4}>80$ but $8$ days do not since $\binom {8}{4}<80$. So the answer is $\text {YES, the murderer can be found in 12 days} $

Nice solution @above Just be careful-Sperner's Lemma is not quite the same as Sperner's Theorem. Interestingly, this is not the first time that this problem has appeared; one version involves theater groups, and is posted somewhere on this forum, although I can't find the link currently.

We can build an explicit construction without Sperner's theorem

Yes just take $80$ different subsets of $\{1,2,3,...,12\}$ each having a cardinality of $6$. Now assign them to the people in an arbitrary order. Let the witness be $w $ and the murderer be $m $. Then both were called on $6$ days and there was a day when $w $ was there and $m $ was not since the assigned subsets are different. So the murderer will be caught. The original problem is simple enough, but I just found it natural to generalise to arbitrary $m,n $ and then solve it.

I have different approach to this problem. Assign a number from $0$ to $79$ to each man, and write these numbers in base $3$. We will have numbers from $0000$ to $2222$. Now we interrogate group that have the same $i-th$ digit, $i$ goes from $1$ to $4$. We will have $12$ different groups and it's guaranteed that in one group we will interrogate the witness without the murderer.