In $n=2$ person (say $A$ & $B$) prisoner's dilemma, the possible outcomes are $AB, CC,CD,DC,DD$ and the payoffs are $(1,1), (0,3), (3,0), (2,2)$ where $C$ is "cooperation" and $D$ is "defection".
Now for $n=3$ $(A,B,C)$, $A$ has two choices (for $B,C$) and the same for $B$ and $C$, therefore the outcomes are
$$\begin{eqnarray*} CC-CC-CC, \\ CC-CC-CD, \\ CC-CC-DC, \\ CC-CC-DD, \\ \dots \end{eqnarray*}$$
and $64$ such others.
What will be the payoff then? I don't intend to get the full payoff but the methodology to calculate the payoff for say $CD-DD-DC$.
PS: I thought of breaking it into $2$ person prisoner's dilemma like the $CD-DD-DC$ as $CD, DD, DC$, evaluate the individual, divide by 3 and sum it up. Is it wrong to do that?
I think you keep the same payoffs, i.e. 1 for everyone if they all co-operate, 2 for everyone if they all defect, 3 if the player defects and isn't defected on and zero otherwise.
Then B defects on both A and C, but both A and C co-operate with B. So we have (0, 3, 0).