expected absolute difference between shuffled range of numbers

If $X_k$ is the $k$-th number after the shuffle then you want to compute

$$ \mathbb{E}\left[\sum_{k=1}^{n-1}|X_{k+1}-X_k|\right]=\sum_{k=1}^{n-1} \mathbb{E}[|X_{k+1}-X_k|]=(n-1)\mathbb{E}[|X_2-X_1|] $$ as $\mathbb{E}[|X_{k+1}-X_k|]$ is the same for any $k \in\{1,\ldots,n-1\}$. Finally, observe that

$$ \begin{align*} \mathbb{E}[|X_2-X_1|]&=\sum_{1\leqslant j,k\leqslant n}|j-k|\Pr [X_1=j]\Pr [X_2=k|X_1=j]\\ &=2\sum_{1\leqslant j<k\leqslant n}(k-j)\frac1{n}\cdot \frac1{n-1}\\ &=\frac{2}{n(n-1)}\sum_{k=2}^n\sum_{j=1}^{k-1}(k-j)\\ &=\frac{2}{n(n-1)}\left(\sum_{k=2}^nk(k-1)-\sum_{k=2}^n\frac{k(k-1)}{2}\right)\\ &=\frac{2}{n(n-1)}\cdot \frac{(n+1)n(n-1)}{6}\\ &=\frac{n+1}{3} \end{align*} $$

Therefore

$$ \mathbb{E}\left[\sum_{k=1}^{n-1}|X_{k+1}-X_k|\right]=\frac{(n-1)(n+1)}{3} $$

Say $X$ is the random variable which represents sum of absolute difference of all $(n-1)$ neighboring pairs.

Leaving aside the order of two numbers in a pair, please notice that there are $(n-i)$ pairs with difference of $~i$
($1 \leq i \leq n-1$).

That means the probability of difference $i$ for a pair of numbers is, $ \displaystyle ( n-i) / {n \choose 2}$. As there are $(n-1)$ neighboring pairs,

$ \displaystyle \mathbb{E}[X] = \left[(n-1) / {n \choose 2}\right] ~ \sum \limits_{i=1}^{n-1} i (n - i) = \frac{2}{n} \sum \limits_{i=1}^{n-1} (n i - i^2)$

As $~\displaystyle \sum\limits_{i=1}^{n-1} i = \frac{n(n-1)}{2}~~ \text {and } ~ \sum\limits_{i=1}^{n-1} i^2 = \frac{n(n-1)(2n-1)}{6}~$,

$ \displaystyle \mathbb{E}[X] = \frac{n^2-1}{3}$