Let $$T(n) = \int_{-\infty}^\infty \frac{e^{-x^2}}{(1+e^{x})^n} dx.$$
We have that $$ T(0) = \sqrt{\pi} \text{ and } T(1) = \frac{\sqrt{\pi}}{2}$$ and also that $$ T(3) = \tfrac{3}{2} T(2) - \frac{\sqrt{\pi}}{4}.$$ Can we find a closed form for $T(2)$? It would also give us $T(3)$. Perhaps something in terms of special functions?
An expression for $T(2)$ as a series of Bernoulli numbers: \begin{align} T(2) &= \int_{-\infty}^\infty \frac{e^{-x^2}}{(1+e^{x})^2} d\,dx\\ &=\frac{1}{4}\int_{-\infty}^\infty \frac{e^{-x^2-x}}{\cosh^2 \frac{x}{2}} \,dx\\ &=\frac{1}{2}\int_{-\infty}^\infty \frac{e^{-4x^2-2x}}{\cosh^2 x} \,dx\\ &=\frac{1}{2}\int_{-\infty}^\infty \frac{e^{-4x^2+2x}}{\cosh^2 x} \,dx \end{align} Summing the last two expressions and using the formula $\cosh 2x=2\cosh^2x-1$, it comes \begin{align} T(2)&=\frac{1}{2}\int_{-\infty}^\infty \frac{\cosh 2x}{\cosh^2 x}e^{-4x^2}\,dx\\ &=\int_{-\infty}^\infty e^{-4x^2}\,dx-\frac{1}{2}\int_{-\infty}^\infty \frac{e^{-4x^2}}{\cosh^2 x}\,dx\\ &=\frac{\sqrt{\pi}}{2}-\frac{1}{2}\int_{-\infty}^\infty \frac{e^{-4x^2}}{\cosh^2 x}\,dx \end{align} We integrate by parts, \begin{equation} T(2)=\frac{\sqrt{\pi}}{2}-4\int_{-\infty}^\infty xe^{-4x^2}\tanh x\,dx \end{equation} Now, using the series expansion for $\tanh$ DLMF: \begin{equation} \tanh z=\sum_{n=1}^\infty\frac{2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}z^{2n-1} \end{equation} one can express, \begin{align} T(2)&=\frac{\sqrt{\pi}}{2}-4\sum_{n=1}^\infty\frac{2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}\int_{-\infty}^\infty x^{2n}e^{-4x^2}\,dx\\ &=\frac{\sqrt{\pi}}{2}-2\sqrt{\pi}\sum_{n=1}^\infty\frac{2^{2n}-1}{2^{2n}}\frac{B_{2n}}{n!} \end{align} where we have used the tabulated formula \begin{equation} \int_0^\infty x^{2n}e^{-px^2}\,dx=\frac{(2n-1)!!}{2(2p)^n}\sqrt{\frac{\pi}{p}} \end{equation}