The tedious way is to expand $(x_1 - x_2)^2 (x_2 - x_3)^2 (x_1 - x_3)^2$ out completely and then write it in terms of $x_1 + x_2 + x_3$ and so forth. You are guaranteed that this is possible by the fundamental theorem of symmetric polynomials, the proof of which even gives an algorithm for doing this, but it's a pain to do by hand (although it's not a bad exercise in algebraic manipulation).
A less tedious way is to argue as follows. We will first work under the assumption that $p = 0$. Now, $q, r$ are polynomials of degrees $2, 3$, and the discriminant is a polynomial of degree $6$, so the discriminant must be a linear combination of the two monomials $q^3, r^2$. Thus we can write
$$\Delta = a q^3 + b r^2$$
for two constants $a, b$, where $\Delta$ is the discriminant. Setting $q = -1, r = 0$ we get the polynomial $x^3 - x = 0$ with roots $0, 1, -1$. We compute that the discriminant is equal to $4$, from which it follows that $a = -4$.
Setting $q = 0, r = -1$ we get the polynomial $x^3 - 1 = 0$ with roots $1, \omega, \omega^2$ where $\omega$ is a primitive third root of unity. Using the identity
$$(\omega - 1)^2 = \omega^2 - 2 \omega + 1 = - 3 \omega$$
we compute that the discriminant is equal to $-27$, from which it follows that $b = -27$. Thus
$$\Delta = -4 q^3 - 27 r^2.$$
To get from here to an arbitrary choice of $x_1, x_2, x_3$, apply the above formula to the polynomial with roots $x_1 - \frac{p}{3}, x_2 - \frac{p}{3}, x_3 - \frac{p}{3}$ and note that subtracting the same constant from each of the three roots doesn't change the discriminant.
For $ k = 0, 1, \ldots, n-1$, consider the (horizontal) vector $v_k$ with $i$th coordinate $$ \sum \prod_{j=1, a_j \neq i}^{k} {x_{a_j}}.$$
For example, with $n = 3$, we have
$v_0 = (1, 1, 1)$,
$v_1 = (x_2 + x_3, x_3 + x_1, x_1 + x_2)$,
$v_2 = ( x_2x_3, x_3x_1, x_1x_2)$.
With $n = 4$, we have
$v_0 = (1, 1, 1, 1)$,
$v_1 = (x_2 + x_3 + x_4, x_3 + x_4 + x_1, x_4 + x_1 + x_2, x_1 + x_2 + x_3)$,
$v_2 = ( x_2x_3+x_3x_4+x_4x_2, x_3x_4+x_4x_1+x_1x_3, x_4x_1+x_1x_2+x_2x_4, x_1x_2 + x_2x_3 + x_3x_2)$,
$v_3 = (x_2x_3x_4, x_3x_4x_1, x_4x_1x_2, x_1x_2x_3)$.
Claim: $v_kA = (n-1-k) v_k$.
Proof: Expand it. A lot of the cross terms cancel out.
For example, with $v_0$, the column sum is $n-1$, so $v_0 A = (n-1) v_0$.
For example, with $v_{k-1}$, the numerators are all $\prod x_i$, and by looking at the denomninators, they cancel out to 0, so $v_{k-1} A = 0 $.
Do you see how we get $v_k A = (n-1-k)v_k$?
Corollary: The eigenvalues are $0, 1, 2, \ldots, n-1 $.
Best Answer
It looks to me like the $12x_1x_2x_3$ term is correct and might be a typo in your source (it's symmetric so you should be able to proceed).
Note that since $\omega$ is a primitive cube root of unity its minimal polynomial is $\omega^2+\omega+1 = 0$, so $3\omega+3\omega^2 = -3$, as desired. The identity $\omega^2+\omega+1 = 0$ also tells us that the sum of the coefficients of $u_1, u_2$ are zero, so we know that the same is true for $u_1^3+u_2^3$, further evidence for the $12x_1x_2x_3$ term being correct.