I guess the main reason why you could not find anything about this is that it is rarely needed (and can be reduced to the better-known boolean operations, as you noted). Checking language equivalence of two automata is close, but there you don't necessarily want to compute the symmetric difference exactly, just check whether it is empty.
One way of improving on the complicated expression is by computing the product of the automata as you would for intersection, but with an accepting condition given by the XOR of those of the original automata. That is, if your automata are $P_1,\dots,P_n$, where $P_i=(S_i,s_i^0,F_i,\tau_i)$ with sets of states $S_i$, initial states $s_i^0$, sets of final states $F_i$ and transition functions $\tau_i:S_i\times A\to S_i$ (assuming the same alphabet $A$ for simplicity), your product is $(S, s^0, F, \tau)$, where
- $S=S_1\times\dots\times S_n$,
- $F$ contains $(s_1,\dots,s_n)$ iff the number of $i$ with $s_i\in F_i$ is odd,
- $\tau((s_1,\dots,s_n),a) = (\tau_1(s_1,a),\dots,\tau_n(s_n,a))$.
The only thing which differs from the usual product is the acceptance condition, reflecting the boolean combination of languages you want (here XOR instead of conjunction).