For the proper method you only invoke the born rule at the very end just in case the SG machines are super sensitive and preserve a well defined phase between the outcomes. But in your case it won't affect the answers to invoke it at every device. So you could imagine starting with 2,4,8, 16, 32, or 64 particles, whichever makes your calculations easier. Send them through your machines each time, have the correct fraction come out each side.
There is an easy way where you only use the Born rule to find out how much is in C, then how much is in A, and then note that the rest end up in B. However, that trick won't work for more complicated setups, so we can sue that as a check again the full solution (however if you haven't yet learned the full method, the following might seem confusing).
Start by writing $|X_+\rangle$ as a linear combination of $|Z_+\rangle$ and $|Z_-\rangle$, then do the same for $|X_-\rangle$. Then repeat again to write $|Z_+\rangle$ as a linear combination of $|X_+\rangle$ and $|X_-\rangle$, then do the same for $|Z_-\rangle$.
Now you are ready. Whenever something enters a SG machine write it as a linear combination of the states associated with the machine, then send each part of the sum to the different branches. So in general the scalars in front of each would get smaller as you branched more and more, but in one case you fed two results into the same machine. The result then depends on the exact speed and length of the paths, but in this case you are sending X states into an X SG so it just sends them through.
How to compute with the Born rule. For numbers $|a+b|^2\neq|a|^2+|b|^2$ but for vectors, they can ... if they are orthogonal. After a strong measurement, there are a bunch of distinct orthogonal outcomes. You can find the squared length of each of them, which is the probability of getting each distinct outcome, and it is all about whether they are distinct (i.e. orthogonal). The $|X_+\rangle$ and $|X_-\rangle$ are orthogonal, and distinct.
Each time you send a particle in you started with a state $|+000\rangle$ where the + says that you have an $|X_+\rangle$ and the three zeros say that each detector is in the ready state. After the machines you end up with a combination like $a|+100\rangle$+$b|+010\rangle$+$c|-010\rangle$+$d|-001\rangle$ where you get the a,b,c, and d from the steps I hinted at in the third and fourth paragraph. The two middle ones both describe the middle detector firing, but they are orthogonal states (since $|X_+\rangle$ and $|X_-\rangle$ are orthogonal), so the sum of the squares is the square of the sum.