We know that a probability of $5/32$ is possible (with the strategy "if you see two reds, guess blue, otherwise guess red"), so let's prove they can't do better.
Suppose there was a strategy with 6 hat combinations $c_1,\dots,c_6$ for which the players win. Let's just look at $c_1,\dots,c_5$ for now.
Connect two combinations with an edge labeled $i$ if player $i$ sees the same thing in both of them. Then two patterns which are connected agree in 3 places: the two places that player $i$ sees, and player $i$'s hat, since what he guesses is determined by what he sees.
Then there must be at least one edge of each label, since player $i$ can only see 4 possible things, and there are 5 patterns, so there must be two where player $i$ sees the same thing.
If two combinations agree on 4 places, they agree on the fifth, since the last hat is forced by surrounding ones.
There can't be more than one edge between any pair of combinations. Otherwise, they would agree in four places, and differ in the fifth, which is impossible by (2).
There are at most two edges out of each combination. If there were 3, there would have to be to edges labeled by adjacent numbers, $i$ and $i+1$; without loss of generality, $i=1$, so we have the below scenario:
$$
c_1\stackrel{1}{\longleftrightarrow}c_2 \stackrel{2}{\longleftrightarrow} c_3
$$
This means $c_2$ and $c_3$ have the same hats on players $1,2,3$. If $c_2$ and $c_3$ also were the same for player 4, then by (2) they would be the same; since $c_2\neq c_3$, they must differ on 4. Similarly, $c_2$ and $c_3$ differ on 4, so it must true that $c_1$ and $c_3$ are the same on 4 (they are opposites of the same thing). So $c_1$ and $c_3$ agree on $1,2$ and $4$, which means they agree on all, which contradicts (3).
Combining (1), (3) and (4) proves that the graph is a 5 cycle, where the edges appearing in the order $1,3,5,2,4$. Thus, $c_1$ uniquely determines $c_2$ through $c_5$. But the same reasoning applied to $c_1,c_2,c_3,c_4,c_6$ shows that $c_6$ must be one of $c_1,\dots,c_5$, so the six combinations were not distinct.