If we let $\triangle$ denote the symmetric difference of two sets, $A\triangle B = (A\cup B) - (A\cap B) = (A-B)\cup(B-A),$ then we have:
Theorem. If $A_1,\ldots,A_n$ are sets, then $\begin{align*} x\in (A_1\triangle\cdots\triangle A_{n-1})\triangle A_n&\iff x\text{ is in exactly an odd number of }A_i\\ &\iff \Bigl|\{i\mid 1\leq i\leq n, x\in A_i\}\Bigr|\text{ is odd.} \end{align*}$
Proof. We proceed by induction on $n$. The result is true if $n=1$ or $2$. Assume the result holds for $k$. Then $x\in (A_1\triangle \cdots\triangle A_n)\triangle A_{n+1}$ if and only if $x$ is in exactly one of $A_1\triangle\cdots\triangle A_n$ and $A_{n+1}$.
If $x$ lies in an even number of sets from among $A_1,\ldots,A_{n+1}$, then it either lies in an even number of sets from among $A_1,\ldots,A_n$ and not in $A_{n+1}$, in which case it lies in neither $A_1\triangle\cdots\triangle A_n$ (by the induction hypothesis) nor in $A_{n+1}$, hence not in the symmetric difference; or else it lies in an odd number of sets from among $A_1,\ldots,A_n$ (and hence lies in $A_1\triangle\cdots\triangle A_n$ by the induction hypothesis) and in $A_{n+1}$, and so it does not lie in the symmetric difference (since it lies in both operands).
If $x$ lies in an odd number of sets from among $A_1,\ldots,A_{n+1}$, then it either lies in an even number of sets from among $A_1,\ldots,A_n$ (and hence not in $A_1\triangle\cdots\triangle A_n$), and in $A_{n+1}$; or it lies in an odd number of sets from among $A_1,\ldots,A_n$ and not in $A_{n+1}$. Either way, it lies in exactly one of $A_1\triangle\cdots\triangle A_n$ and $A_{n+1}$, hence lies in $(A_1\triangle\cdots\triangle A_n)\triangle A_{n+1}$.
Thus, $x\in (A_1\triangle\cdots\triangle A_n)\triangle A_{n+1}$ if and only if it lies in exactly an odd number of sets from among $A_1,\ldots,A_{n+1}$. $\Box$
Moreover, since $(A\triangle B)\triangle C = A\triangle(B\triangle C)$ for all $A$, $B$, and $C$ (it follows easily from the theorem, or from truth tables), we can dispense with the parentheses.
Essentially, the characteristic function of the symmetric difference is the sum, modulo $2$, of the characteristic functions of the two sets: $1_{A\triangle B} = 1_A\oplus 1_B,\qquad \oplus=\text{addition modulo }2.$