I suppose I should make the mathematics a bit more clear:
The state of the system is ##\left|\psi\right> = (\left|\psi_1\right>\otimes \left|\psi_2\right> - \left|\psi_2\right> \otimes \left|\psi_1\right>) \otimes (\left|\psi_3\right>\otimes \left|\psi_4\right> - \left|\psi_4\right> \otimes \left|\psi_3\right>)##. You can now do two things:
1. Compute the reduced density matrix in ##\mathcal H_1 \otimes \mathcal H_4## and call it ##\rho_\text{no measurement}##.
2. Perform the Bell state measurement on particles 2 & 3. In that case, you get 4 orthogonal projectors ##P_1,\ldots P_4## and thus 4 states ##\left|\xi_i\right> = P_i \left|\psi\right>##. Now you can compute the density matrix ##\rho_\text{BSM} = \sum_i \left|\xi_i\right>\left<\xi_i\right|## and again compute the reduced density matrix in ##\mathcal H_1 \otimes \mathcal H_4## and call it ##\rho_\text{measurement}##.
Of course, you have to add all the normalization constants, which I omitted for brevity. You will find that mathematically ##\rho_\text{no measurement} = \rho_\text{measurement}##. This shows that the Bell state measurement on particles 2 & 3 can not influence the physical state of the composite 1 & 4 system. Of course, this does not prevent us from performing entanglement swapping on the recorded data.I did view the experiment in the total context. The question is: Can the actions on particles 2 & 3 have a causal influence on the physical state of the 1 & 4 system? For this to be true, the state of the 1 & 4 system must change depending on what one does to the 2 & 3 system, but it doesn't. Yes, the situation is very analogous to the quantum eraser. It's also an effect of post-selection, which can be understood from standard quantum theory. The experiment says nothing about causality. It's just yet another indication that the predictions of quantum mechanics are correct.