A sort of (partial) explanation for what happens:

Let $N \ge 3$ the degree and let $A=\sum{a_k}, B=\sum_{j<k}a_ja_k, C=\sum_{j\ne k\ne m \ne j}a_ja_ka_m $. We are given that $A \ge 0, C \ge 0$ and we need to prove that $N^2(A^3-3AB+3C) \ge A^3$. Now we can assume wlog $A =1$ since if $A=0$ the inequality is obvious and otherwise we can divide by $A>0$ and consider $c_j=\frac{a_j}{A}$ and prove the inequality for them etc

So we need to prove $1-3B+3C \ge \frac{1}{N^2}$ under the hypothesis as above (the polynomial $X^N-X^{N-1}+BX^{N-2}-CX^{N-3}+...$ has real roots and $C \ge 0$

Then if we let $b_j=a_j-\frac{1}{N}, A_1,B_1,C_1$ the corresponding symmetric polynomials in $b_j$ we have $A_1=0, B_1=B-\frac{N-1}{2N}=B-\frac {1}{2}+\frac{1}{2N}, C_1=C-B+\frac{2B}{N}+\frac{1}{3}-\frac{1}{N}+\frac{2}{3N^2}$ so the inequality becomes ($C-B=C_1+...$ from the last equality)

$1+3C_1-\frac{6B}{N}-1+\frac{3}{N}-\frac{2}{N^2}\ge \frac{1}{N^2}$ and since

$\frac{6B}{N}=\frac{6B_1}{N}+\frac{3}{N}-\frac{3}{N^2}$ all reduces to

$3C_1-\frac{6B_1}{N} \ge 0$

But now the polynomial $X^N+B_1X^{N-2}-C_1X^{N-3}+...$ has real roots too as they are just $b_k$ and hence $B_1 \le 0, B_1=-B_2, B_2 \ge 0$ so the inequality reduces to $2B_2+NC_1 \ge 0$ and we know that $C_1=C+\frac{N-2}{N}B_2-\frac{(N-1)(N-2)}{6N^2}$

So we need $C_1$ negative but $C \ge 0$

By differentiating $N-3$ times and using Gauss Lucas/Rolle (so the cubic that results which is in standard form) has real roots so $4(-p)^3 \ge 27q^2$, we get some constraints on $B_2, -C_1$ which are enough to give the result for $N \le 6$ with some crude approximations

Then if we try easy counterexamples for the $b_k$ of the type $N-1$ $a$ and one $-(N-1)a$ we can solve at $N=10$, $a > \frac{3}{80}$ close enough to it to satisfy the inequality $C>0$ (which is satisifed at $a=\frac{3}{80}$ that one giving equality in the OP inequality as normalizing to integers we get $11$ taken $9$ times, $-19$ taken once and it is easy to see that the constraints are good and $S_1=80, S_3=5120$ and obviously $100\cdot 5120=80^3$

So as noted in the comments taking $9$ of $111$ and one of $-199$ gets a counterexample with a positive sum of cubes (corresponding to $a=\frac{3}{80}+\frac{1}{800}$ normalized to integers)

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