This is problem 3.3.3.(b) in Probability and Random Processes by Grimmett and Stirzaker.
Here's my attempted solution:
We introduce the random variables $\{X_{ij}\}$, denoting the scores of each pair (player $i$ and player $j$), and the total score $Y = \sum_{i<j}X_{ij}$. We calculate the expected value of $Y$: $$ \mathbb{E}(Y) = \sum_{i<j}\mathbb{E}(X_{ij}) = {n\choose{2}}\mathbb{E}(X_{12}) = \frac{7}{12}{n\choose{2}}. $$ Now, let's determine the variance of $Y$: $$ \mathrm{var}(Y) = \mathbb{E}(Y^2) - \mathbb{E}(Y)^2 = \mathbb{E} \left\{ \left( \sum_{i<j}X_{ij} \right)^2 \right\} - \mathbb{E}(Y)^2 = \mathbb{E} \left( \sum_{i<j,\space k<l}X_{ij}X_{kl} \right) - \mathbb{E}(Y)^2 . $$ Further, we look closer at the sum $\sum_{i<j,\space k<l}X_{ij}X_{kl}$. Here $X_{ij}$ and $X_{kl}$ are independent whenever $i\neq k$, $i\neq l$, $j\neq k$ and $j\neq l$.
When both $i = k$ and $j = l$ we get the random variables $\{X_{ij}^2\}_{i<j}$, each having expected value $$ \mathbb{E}(X_{ij}^2) = \mathbb{E}(X_{12}^2) = \sum_{m=1}^6\frac{m^2}{36}=\frac{91}{36}. $$ When only three of the four inequalities above hold we get a random variable on one of the following forms: $X_{ij}X_{jl}$, $X_{ij}X_{il}$, $X_{ij}X_{ki}$ or $X_{ij}X_{kj}$. These have expected value $$ \mathbb{E}(X_{ij}X_{il}) = \mathbb{E}(X_{12}X_{13}) = \sum_{m=1}^6\frac{m^2}{216}=\frac{91}{216}. $$ We note that each triple $\{a,b,c\}$, such that $1\leq a < b < c \leq n$, is associated to the following six terms $X_{ab}X_{ac}$, $X_{ac}X_{ab}$, $X_{ab}X_{bc}$, $X_{bc}X_{ab}$, $X_{ac}X_{bc}$, $X_{bc}X_{ac}$, all with the above expected value. Clearly there are ${n\choose{3}}$ such triples.
This gives us the following: $$ \mathrm{var}(Y) = \mathbb{E} \left( \sum_{i<j,\space k<l}X_{ij}X_{kl} \right) - \mathbb{E}(Y)^2 $$ $$ = {n\choose{2}}\mathbb{E}(X_{12}^2) + 6{n\choose{3}}\mathbb{E}(X_{12}X_{13})+ \left\{{n\choose{2}}^2 - {n\choose{2}} - 6{n\choose{3}} \right\}\mathbb{E}(X_{12})^2 - \left\{{n\choose{2}}\mathbb{E}(X_{12})\right\}^2 $$ $$ = \frac{35}{16}{n\choose{2}} + \frac{35}{72}{n\choose{3}} $$
The book of solutions gives the answer $\frac{35}{16}{n\choose{2}} + \frac{35}{432}{n\choose{3}}$ though, which is what I would get if each triple $\{a,b,c\}$ only was associated with one term $X_{ab}X_{bc}$. So I guess I'm asking why(/if) that is the case!
Your solution appears to be correct. If $n=3$, there are $216$ equally likely outcomes of the dice. They're easy to enumerate, and the variance of the scores comes out to be $\frac{1015}{144}$. This agrees with your answer, not the book's.