A minor extension of the dining cryptographers protocol ( a b )
can be used to reveal 1 bit of information: "all 3 keys are identical", or "at least 1 of the keys is different".
(If only 2 are identical, and the third is different,
this protocol only indicates "at least 1 of the keys is different".
So this doesn't completely answer your question, since it seems you want to reveal more information in that case).
4 persons A, B, C, and D sit around a circular table; each one knows the secret numbers sA, sB, sC, and sD.
There are also 4 fair coins on the table, halfway between each pair of neighbors: cAB, cBC, cCD, cDA.
One round goes something like this:
Each person generates a fresh new random number and announces it everyone else.
Everyone combines the 4 new random numbers to generate a common random number R = rA xor rB xor rC xor rD.
Each person uses some cryptographic hash function to deterministically produce 1 bit hX from their secret number and the common random number,
perhaps hA = least_significant_bit_of( SHA256( sA concatenate R ) ).
Each pair of neighbors secretly chooses a bit at random.
In other words, A and B use the coin cAB between them to choose a bit at random,
not telling anyone else what that bit is,
and so on for the other 3 coins.
Each person calculates a message mX from their hX value and the bits from the coins to their left and right.
In particular, D calculates mD = cCD xor sD xor cDA.
Then everyone announces their messages,
and then computes the xor of all 4 announced messages:
M = mA xor mB xor mC xor mD.
If everyone's secret value is the same, then M will always be zero for every round.
So as soon as M is calculated to be 1, everyone knows that "at least 1 of the keys is different".
If everyone's secret value is the different, then about half the time M will be 1 on the first round.
And if everyone's secret value is different and they find M is 0 on the first round, then about half the time M will be 1 on the second round.
So you have to run through several times to gain confidence that the values really are the same.
The above protocol only directly applies when there is an even number of people around the table.
If there is an odd number of people, then you could have each person pretend to be 2 people in the above protocol -- 3 people could pretend to be 6 people in the above protocol.