= Solution
Let $T$ swap coordinate blocks $0,\ldots,n$ and $n+1,\ldots,2n+1$, fixing all other coordinates. The symmetric law makes $T$ measure preserving, and invariance of the event $A$ gives $F\circ T=F$ pointwise. Put $\mathcal B_n=\sigma(X_{n+1},\ldots,X_{2n+1})$.
Conditional expectation transforms covariantly under this block swap:
$$
\mathbb E[F\mid\mathcal E_n]\circ T=\mathbb E[F\mid\mathcal B_n]\quad\text{almost surely}.
$$
To check it, the left side is $\mathcal B_n$-measurable, and for $C\in\mathcal B_n$ the change of variables by $T$ sends its defining integral to the conditional-expectation identity on $T(C)\in\mathcal E_n$, with $F\circ T=F$. Thus the two versions agree. Measure preservation now gives
$$
\boxed{\|F-\mathbb E[F\mid\mathcal B_n]\|_1
=\|F-\mathbb E[F\mid\mathcal E_n]\|_1\leq\varepsilon.}
$$
This is the first requested estimate.
The <L1 contraction of conditional expectation> follows from the pointwise bound $|\mathbb E[Z\mid\mathcal H]|\leq\mathbb E[|Z|\mid\mathcal H]$: take expectations to get $\|\mathbb E[Z\mid\mathcal H]\|_1\leq\|Z\|_1$. Applied to a difference, it makes conditional expectation a contraction between any two integrable inputs.
Let $U_n=\mathbb E[F\mid\mathcal B_n]$. Since $\mathcal B_n\subseteq\mathcal G_n$, $U_n$ is $\mathcal G_n$-measurable. Hence
$$
\|F-\mathbb E[F\mid\mathcal G_n]\|_1
\leq\|F-U_n\|_1+\|\mathbb E[U_n-F\mid\mathcal G_n]\|_1
\leq2\varepsilon.
$$
Combining this with the other given approximation bound yields
$$
\|F-\mathbb E[F\mid\mathcal G]\|_1\leq3\varepsilon.
$$
The left side does not depend on $n$, and $\varepsilon>0$ was arbitrary. It is therefore zero. For a $\mathcal G$-measurable version $Y=\mathbb E[F\mid\mathcal G]$, the event $C=\{Y>1/2\}$ belongs to $\mathcal G$ and satisfies
$$
\boxed{\mathbb P(A\mathbin{\triangle}C)=0.}
$$
This proves that <exchangeable invariant events are tail events modulo null sets>. It does not assert literal pointwise equality of the two events or sigma-algebras before completing them with null sets.
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