Suppose $\{\mathbf{X}_1,\mathbf{X}_2,\cdots,\mathbf{X}_n\}$ are $n$ iid random variables uniformly drawn from $\{-1,1\}$ and $n$ is odd. Then \begin{align} \mathbb{E}|\sum_{i=1}^n\mathbf{X}_i|= 2\sum_{k=0}^{(n-1)/2} {n\choose k}(2n-k)2^{-n}=n{n-1\choose (n-1)/2}2^{1-n} \end{align} The above can be verified by some calculations.
Why does it hold that every $\mathbf{X}_i$'s contribution to $\mathbb{E}|\sum_{i=1}^n\mathbf{X}_i|$ is the probability that exactly half of the other random variables are taken $1$ and half are taken $-1$? This can be seen from simply do some calculations but are there any intuitive explanations?
The additional $X_i$ is going to boost the expectation of the absolute value of the sum if it's got the same sign as the rest of the sum, and is going to decrease that quantity if it's got the opposite sign; these two cases (same/opposite sign) are equiprobable and therefore the contribution from $X_i$ in these two cases cancels out.
That leaves the case that the other summands exactly cancel, in which case $X_i$ always contributes exactly 1.