How to evaluate $\mathbb{E}(|X_1-X_2|)?$

The definition seems more helpful here.

Denote our sample space here with $\Omega = \{(i,j):1\leq i,j \leq 6\}$ where the first coordinate gives the value first dice and the second coordinate gives the value of the second dice.

Let $X_1, X_2$ as you defined and put $X= \max\{X_1, X_2\}$. Then $X$ attains the values $1,2,3,4,5,6$ and you must calculate

$$\mathbb{E}[X] = \sum_{i=1}^6 i\mathbb{P}(X = i)$$

Thus $$\mathbb{P}(X=1) = \mathbb{P}(\{(1,1)\}) =1/36$$ $$\mathbb{P}(X = 2) = \mathbb{P}(\{(1,2),(2,2),(1,2)\}=3/36$$ $$\mathbb{P}(X = 3) = \mathbb{P}(\{(1,3),(1,3),(2,3), (3,2), (3,3)\}=5/36$$ $$\mathbb{P}(X = 4) = \mathbb{P}(\{(1,4),(4,1),(2,4), (4,2), (3,4), (4,3), (4,4)\}=7/36$$ $$\mathbb{P}(X=5) = \dots = 9/36$$ $$\mathbb{P}(X=6) = \dots =11/36$$

and we get

$$\mathbb{E}[X]= 1/36 + 2.3/36 + 3.5/36 + 4.7/36 + 5.9/36 + 6. 11/36 = 161/36$$

Of course, you can still evaluate $\mathbb{E}[|X_1-X_2|]$ but calculating explicit probabilities cannot be avoided here.

We can easily use conditioning (that's the golden rule of using expectation, whenever stuck in a deadend, use conditioning) $$\mathbb{E}[|X_1-X_2|]=\mathbb{E}[X_1-X_2 \mid X_1 \ge X_2] + \mathbb{E}[X_2-X_1 \mid X_2 > X_1]$$ Now notice that if $X_1 \ge X_2$ then $X_1-X_2 =0,1,2,3,4,5$ with probabiilities $\frac{6}{36},\frac{5}{36},\frac{4}{36},\frac{3}{36},\frac{2}{36},\frac{1}{36}$, respectively.

If $X_2 \ge X_1$ then $X_2-X_1 =1,2,3,4,5$ with probabiilities $\frac{5}{36},\frac{4}{36},\frac{3}{36},\frac{2}{36},\frac{1}{36}$, respectively.

Thus we have

$$\mathbb{E}[X_1-X_2 \mid X_1 \ge X_2]= \frac{6}{36}(0)+\frac{5}{36}(1)+\frac{4}{36}(2)+\frac{3}{36}(3)+\frac{4}{36}(2)+\frac{1}{36}(5) = \frac{35}{36}$$ $$\mathbb{E}[X_2-X_1 \mid X_2 > X_1]=\frac{5}{36}(1)+\frac{4}{36}(2)+\frac{3}{36}(3)+\frac{4}{36}(2)+\frac{1}{36}(5) = \frac{35}{36}$$

Alongside your own part of answer we have $\frac{1}{2}[7+\frac{70}{36}]=\frac{161}{36}$

Another way is to evaluate the $\mathbb{E}[|X_1-X_2|]$ directly. We know that $\mathbb{E}[|X_1-X_2|] = 0,1,2,3,4,5$ with probabilities $\frac{6}{36},\frac{10}{36},\frac{8}{36},\frac{6}{36},\frac{4}{36},\frac{2}{36}$, respectively. Thereforewe have $$\mathbb{E}[|X_1-X_2|]= \frac{6}{36}(0)+\frac{10}{36}(1)+\frac{8}{36}(2)+\frac{6}{36}(3)+\frac{4}{36}(4)+\frac{2}{36}(5) = \frac{70}{36}$$

Your argument is correct. Indeed

$$ \mathbb{E}[| X_1 - X_2| ] =\frac{1}{36} \sum_{x_1=1}^6 \sum_{x_2=1}^6 |x_1-x_2| = \frac{35}{18} $$

And $$ \frac{1}{2} \left ( 7 + \frac{35}{18} \right ) = \frac{161}{36} $$

Actually, computing $\mathbb E\max(X_1,X_2)$ and $\mathbb E|X_1-X_2|$ are equally difficult, so I find it clearest to just compute $\mathbb E\max(X_1,X_2)$ directly as follows, and for $n$-sided die instead of $6$: $$ \mathbb E\max(X_1,X_2)=\frac{1}{n^2}\sum_{i=1}^n\sum_{j=1}^n\max(i,j). $$ Now we can split the sum into three parts: terms with $i<j$, terms with $i=j$, and terms with $i>j$. By symmetry the first and last are the same, so we get $$ \mathbb E\max(X_1,X_2)=\frac{1}{n^2}\left(2\sum_{i=1}^ni(i-1)+\sum_{i=1}^ni\right). $$ Fortunately both sums have simple closed forms: $$ \sum_{i=1}^ni=\frac{n^2+n}{2},\qquad \sum_{i=1}^ni(i-1)=\frac{n^3-n}{3}, $$ both of which can be verified by induction (or in other slicker ways...)

Plugging in gives the result $$ \mathbb E\max(X_1,X_2)=\frac{4n^2+3n-1}{6n}. $$