= Solution
Let the summands be <independent and identically distributed random variables> with mean $\mu$, <variance> $\sigma^2>0$, and standardized third <cumulant> $\gamma_1=\mathbb E[(Y_1-\mu)^3]/\sigma^3$. Define
$$
S_n^*=\frac{\sum_{i=1}^n(Y_i-\mu)}{\sigma\sqrt n}.
$$
An <Edgeworth expansion> refines the <central limit theorem> using the higher <cumulants> of one summand. Some smoothness and moment assumptions are essential: finite <variance> alone does not supply a density expansion. A convenient sufficient set for the expansions here is a smooth density whose <derivatives> decay faster than every inverse power, together with $\mathbb E e^{\delta|Y_1|}<\infty$ for some $\delta>0$. These assumptions give all required moments, Fourier smoothing and the nonlattice characteristic-function condition. Less restrictive sufficient conditions are possible, but this set covers the logistic model below.
Write $\phi$ and $\Phi$ for the <standard normal density> and <standard normal distribution function>. The first-order density <Edgeworth expansion> is
$$
\boxed{p_{S_n^*}(x)=\phi(x)\left[1+\frac{\gamma_1}{6\sqrt n}(x^3-3x)\right]+O(n^{-1}).}
$$
The remainder is uniform on every fixed compact interval under the stated assumptions. To see the origin of the correction, for $Z=(Y_1-\mu)/\sigma$ the <cumulant expansion> of its <characteristic function> gives
$$
n\log\mathbb E e^{itZ/\sqrt n}
=-\frac{t^2}{2}+\frac{\gamma_1(it)^3}{6\sqrt n}+O(n^{-1}).
$$
Exponentiating and using <Fourier inversion> yields the displayed correction, since the inverse transform of $(it)^3e^{-t^2/2}$ is $-\phi^{(3)}(x)=(x^3-3x)\phi(x)$. The <Probabilists' Hermite polynomial> convention is being used: $H_2(x)=x^2-1$ and $H_3(x)=x^3-3x$.
The corresponding <distribution function> expansion is
$$
\boxed{F_{S_n^*}(x)=\Phi(x)-\frac{\gamma_1}{6\sqrt n}(x^2-1)\phi(x)+O(n^{-1}),}
$$
uniformly on fixed compact intervals. The sign follows from $\frac{d}{dx}[H_2(x)\phi(x)]=-H_3(x)\phi(x)$. These are expansions of both the density and <distribution function>; a uniform density error alone would not justify integrating an error bound over the entire real line without the accompanying tail control.
For a fixed $\alpha\in(0,1)$, write $z=z_\alpha=\Phi^{-1}(\alpha)$. At leading order the proposed <quantile> expansion gives $\Phi(p_0(z))=\Phi(z)$, and strict monotonicity gives $p_0(z)=z$. Substitute $y_\alpha=z+p_1(z)n^{-1/2}+O(n^{-1})$ into the <distribution function> expansion. A <Taylor expansion> at $z$ gives
$$
F_{S_n^*}(y_\alpha)
=\alpha+\frac{\phi(z)}{\sqrt n}\left[p_1(z)-\frac{\gamma_1}{6}(z^2-1)\right]+O(n^{-1}).
$$
Since $\phi(z)>0$, the coefficient of $n^{-1/2}$ must vanish. Thus the first <Cornish-Fisher expansion> coefficients are
$$
\boxed{p_0(z)=z,\qquad p_1(z)=\frac{\gamma_1}{6}(z^2-1).}
$$
For the <logistic distribution>, set $Z_i=Y_i-\theta$. Its density $q(z)=e^z/(1+e^z)^2$ satisfies $q(-z)=q(z)$ and has exponentially decreasing tails. Symmetry and integrability give $\mathbb E Z_i=0$, and the supplied integral gives
$$
\mathbb E Z_i^2=2\int_0^\infty\frac{z^2e^z}{(1+e^z)^2}\,dz=\frac{\pi^2}{3}.
$$
Therefore $\mathbb E Y_i=\theta$, $\operatorname{Var}(Y_i)=\pi^2/3$, and the <central limit theorem> gives
$$
T_n=\frac{\sqrt n(\bar Y-\theta)}{\pi/\sqrt3}\xrightarrow{d}N(0,1).
$$
Writing $z_-=z_{\alpha/2}$ and $z_+=z_{1-\alpha/2}$, the condition $z_-\leq T_n\leq z_+$ is equivalent to
$$
\bar Y-\frac{\pi}{\sqrt{3n}}z_+\leq\theta\leq
\bar Y-\frac{\pi}{\sqrt{3n}}z_-.
$$
Consequently the <confidence interval>
$$
\boxed{\left(\bar Y-\frac{\pi}{\sqrt{3n}}z_{1-\alpha/2},\quad
\bar Y-\frac{\pi}{\sqrt{3n}}z_{\alpha/2}\right)}
$$
has limiting coverage $\Phi(z_+)-\Phi(z_-)=1-\alpha$. Whether the finite endpoints are included makes no probability difference because the sample mean has a density.
The centered <logistic distribution> is symmetric, so $\gamma_1=0$. Its smooth density and exponentially decreasing <derivatives> satisfy the conditions stated above; in particular all moments exist and Fourier smoothing is available. The first correction in the <Edgeworth expansion> vanishes, leaving $F_{T_n}(z)=\Phi(z)+O(n^{-1})$ at each of the two fixed <quantiles>. Subtracting gives the stronger conclusion
$$
\boxed{P_\theta(\theta\text{ lies in the interval})=1-\alpha+O(n^{-1}).}
$$
The standardized law is independent of $\theta$, so this coverage-error conclusion holds with the same bound for every location parameter.
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