Does partial tracing a system with three shared Bell states give the identity?

The result is correct. You can see it in the other way. You have three two-qubit subsystems $A = {1,2}$, $B = {3,4}$ and $C = {5,6}$. The whole state is the product state on those three subsystems, i.e. the whole density matrix is $rho_A otimes rho_B otimes rho_C$. Product state means there are absolutely no correlations between the states on subsystems. Hence you can fully ignore the $C$ subsystem if you are interested only in the state on $D = {1,4}$. Also since $rho_{AB} = rho_{A} otimes rho_{B}$ then it must be $rho_D = rho_1 otimes rho_4$.

What does this density matrix say about the information that the players have about each others particles?

Nothing, if no further assumptions are made on the initial state. If $A$ and $B$ share a state $rho$, a reduced state $rho^A$ doesn't say anything about the information that $A$ has about $B$'s state. Indeed, $B$ can have any state as far as $A$ knows: any shared state of the form $rho^Aotimes sigma$ is compatible with $A$'s knowledge, for any state $sigma$.

If on the other hand Alice knows that the initial state is pure, then knowing that their share of the state is $rho^A$, $A$ can infer the entropy of $B$'s state, because $S(rho^A)=S(rho^B)$. Nothing else can be said though, as any local operation applied to $B$ will not change $A$'s outcomes.