Answer: 132

Solution:

The roots of the cubic are $\alpha^2, \beta^2, \gamma^2.$ Fix square roots $s_1, s_2, s_3$ of them; then $\alpha + \beta + \gamma$ ranges over the eight expressions $\pm s_1 \pm s_2 \pm s_3,$ which come in four pairs $\pm v.$ Generically all eight are distinct. Their product is nonzero because the cubic's constant term is $-400,$ so every $s_i$ is nonzero. A coincidence $v(\varepsilon) = v(\varepsilon')$ between choices that are not opposite must differ in two signs and forces $s_i = \pm s_j$ for some $i \ne j,$ which collapses the eight values to at most six. So exactly seven values occur precisely when one choice satisfies $\pm s_1 \pm s_2 \pm s_3 = 0$ — its opposite is then the same value $0$ — and no further degeneracies occur.

That condition is the vanishing of $$\begin{aligned} &(s_1 + s_2 + s_3)(-s_1 + s_2 + s_3) \\ &\quad {}\cdot (s_1 - s_2 + s_3)(s_1 + s_2 - s_3) \\ &= 2\sum_{i \lt j} r_i r_j - \sum_i r_i^2 \\ &= 4e_2 - e_1^2, \end{aligned}$$ where $r_i = s_i^2$ are the roots and $e_1, e_2$ their elementary symmetric functions. By Vieta's formulas $e_1 = \frac{n}{6}$ and $e_2 = n - 11,$ so $\frac{n^2}{36} = 4(n - 11),$ i.e. $n^2 - 144n + 1584 = 0,$ with roots $n = 12$ and $n = 132.$

For $n = 132$ the cubic factors as $$(x - 16)(x^2 - 6x + 25),$$ with distinct roots $16,$ $3 + 4i,$ and $3 - 4i.$ Choose their square roots as $4,$ $2 + i,$ and $2 - i;$ then $4 - (2+i) - (2-i) = 0.$ The eight sign choices give $0$ twice and the six distinct nonzero values $\pm 8,$ $\pm(4+2i),$ $\pm(4-2i),$ so exactly seven sums occur. The greatest such integer is $132.$