= Solution
Identify a subset $A\subseteq\Omega$ with its indicator <vector> $a=(a_1,\ldots,a_n)\in\mathbb F_2^n$. <Symmetric difference> becomes coordinatewise addition modulo two, so the subsets form a <vector space> of dimension $n$, with the singleton subsets as a <basis>. The proposed form becomes the dot product
$$
B(a,b)=\sum_{i=1}^na_ib_i\in\mathbb F_2.
$$
Distributivity proves bilinearity, and commutativity proves symmetry. Its <Gram matrix> in the singleton <basis> is $I_n$, so it is <nondegenerate>. A <permutation> merely permutes coordinates and preserves the dot product, proving invariance under $S_n$.
Write $u=(1,\ldots,1)$ for the full subset. Its <bilinear orthogonal complement> $E=u^\perp$ consists exactly of even-cardinality subsets. If $n\geq2$ is even, $B(u,u)=n=0$ in $\mathbb F_2$, so $\langle u\rangle\leq E$. For any $v\in E$, $B(v,v)=\sum_iv_i^2=\sum_iv_i=0$, so the restricted form is alternating.
Nondegeneracy of the full form gives $E^\perp=\langle u\rangle$; one can also see this directly by testing against all <vectors> $e_i+e_j$, which forces all coordinates of an element of $E^\perp$ to be equal. Hence the <radical of a bilinear form> of the restricted form is
$$
\operatorname{rad}(B|_E)=E\cap E^\perp=\langle u\rangle.
$$
Define the quotient form by $\overline B(v+\langle u\rangle,w+\langle u\rangle)=B(v,w)$. Adding $u$ to either representative does not change the pairing with $E$, so it is well-defined. It is alternating, and a class pairing to zero with all classes has a representative in $\operatorname{rad}(B|_E)=\langle u\rangle$, hence is zero. Therefore
$$
\boxed{\langle\Omega\rangle^\perp/\langle\Omega\rangle\text{ is a nondegenerate alternating space of dimension }n-2.}
$$
Both spaces are $S_n$-invariant, so the coordinate action descends and preserves the quotient form. This is the <symplectic quotient of the binary subset module>. The evenness assumption is essential because otherwise $u$ does not lie in its own <bilinear orthogonal complement>.
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