UPDATE (2022-07-13). The generating function for $A_k$ can be expressed as
$$\sum_{k\geq0} A_k t^k = {\cal L}_{x_1,\dots,x_n,y_1,\dots,y_n} \sum_{\lambda} e_{\lambda}(x_1,\dots,x_n)\cdot m_{\bar\lambda}(y_1,\dots,y_n)\cdot t^{\mathrm{sum}(\lambda)},$$
where summation is done over all partitions $\lambda$ whose Young tableau fits the $n\times n$ square; $\bar\lambda$ is the partition whose Young tableau complements that of $\lambda$ in the $n\times n$ square; $e$ and $m$ are elementary and monomial symmetric polynomials respectively; and $\cal L$ is the Laplace transform evaluated at $1$, which replaces each $x_i^d$ or $y_j^d$ with $d!$.
With this formula I was able to extend computations and confirm the conjecture for $n\leq 10$. The data is uploaded to OEIS A261602 and OEIS A261603.
Below is my original answer presenting data for $n\leq 6$.
I've computed values of $|A_k|$ for $0\leq k\leq \lfloor n^2/2\rfloor$ and $n\leq 6$:
$n=1:$ 1
$n=2:$ 4, 8, 10
$n=3:$ 216, 648, 1188, 1668, 1944
$n=4:$ 331776, 1327104, 3151872, 5695488, 8608896, 11446272, 13791744, 15326208, 15858432
$n=5:$ 24883200000, 124416000000, 360806400000, 787138560000, 1426595328000, 2262299258880, 3240594432000, 4283587584000, 5304730521600, 6222411878400, 6968709089280, 7493189990400, 7763310604800
$n=6:$ 139314069504000000, 835884417024000000, 2855938424832000000, 7259810955264000000, 15220062093312000000, 27765294052147200000, 45532546213478400000, 68600569724928000000, 96440964380098560000, 127985462154362880000, 161777817980986982400, 196164002436769382400, 229476155622594969600, 260178812386069708800, 286962944406552576000, 308788668410898677760, 324887962565624463360, 334743605500457779200, 338060641751949312000
So the conjecture is confirmed numerically for $n\leq 6$.
You can prove this by using a result of H. B. Mann to reduce to the case that $n$ is squarefree, and then applying the Lam-Leung result mentioned in one of the comments above.
Theorem 1 of Mann's paper "On linear relations between roots of unity" (Mathematika 12 (1965), 107-117) says the following. Let $\zeta_1,\dots,\zeta_r$ be roots of unity and let $a_1,\dots,a_r$ be nonzero integers. Suppose that $\sum_{i=1}^r a_i \zeta_i = 0$ but that there is no nonempty proper subset $S$ of $\{1,2,\dots,r\}$ for which $\sum_{i\in S} a_i \zeta_i = 0$. Then for each $i$ and $j$ we have $(\zeta_i/\zeta_j)^m=1$ where $m$ is the product of the primes which are $\le r$.
To deduce what you want, start with your equality
$$\zeta^{i_1}+\dots+\zeta^{i_k}=\zeta^{j_1}+\dots+\zeta^{j_k}.$$
First remove any terms which appear on both sides of the equation, to get a similar equation but possibly with a smaller value of $k$; I will assume that the resulting equation still contains some roots of unity (which will lead to a contradiction). Next multiply both sides of this last equation by the reciprocal of one of the roots of unity appearing in the equation, in order to obtain a similar equation in which one of the roots of unity equals $1$. As noted in one of the comments above, if every root of unity appearing in this last equation has squarefree order then we get a contradiction from Theorem 4.1 of Lam-Leung's paper "On vanishing sums of roots of unity" (J.Algebra 224 (2000), 91-109). So assume that some root of unity in the equation does not have squarefree order.
Subtract the right side of the equation from the left to get $\sum_{i=1}^r a_i\zeta_i=0$
where $r\le 2k$, each $a_i$ is in $\{1,-1\}$, and $\zeta_1,\dots,\zeta_r$ are distinct $n$-th roots of unity. If there exists a nonempty proper subset $S$ of $\{1,2,\dots,r\}$ for which $\sum_{i\in S} a_i \zeta_i = 0$ then the smallest such $S$ has size at most $r/2$ which is at most $k$, whence by Mann's theorem any distinct $i,j\in S$ satisfy $(\zeta_i/\zeta_j)^m=1$ where $m$ is the product of the primes which are $\le k$. But also $(\zeta_i/\zeta_j)^n=1$ where $n$ is coprime to $m$, so that $\zeta_i/\zeta_j=1$, contrary to our hypothesis that $i,j$ are distinct. Hence there is no such subset $S$, so by Mann's theorem there is a squarefree positive integer $m$ such that all $i,j$ with $1\le i,j\le r$ satisfy $(\zeta_i/\zeta_j)^m=1$. Since some $\zeta_j$ equals $1$, it follows that $\zeta_i^m=1$ for each $i$, contradicting our assumption that some root of unity in the equation does not have squarefree order.
Incidentally, the proof of Mann's theorem is beautiful. Let me just give the first step to convey the flavor. Let $\zeta_1,\dots,\zeta_r$ be roots of unity and let $a_1,\dots,a_r$ be nonzero integers such that $\sum_{i=1}^r a_i \zeta_i = 0$. Let $n$ be the least common multiple of the orders of the $\zeta_i$'s. Suppose that $n$ is divisible by the square of some prime $p$. Let $\zeta$ be a primitive $n$-th root of unity, and write $\zeta_i=\zeta^{c_i}$ for some integer $c_i$. Then write $c_i=pd_i+e_i$ with $d_i,e_i\in\mathbf{Z}$ and $0\le e_i\le p-1$. This yields $\sum_{i=1}^r a_i (\zeta^p)^{d_i} \zeta^{e_i} = 0$, which is a vanishing linear combination of $\zeta^0,\zeta^1,\dots,\zeta^{p-1}$ having coefficients in the field $\mathbf{Q}(\zeta^p)$. But $$[\mathbf{Q}(\zeta):\mathbf{Q}(\zeta^p)]=\frac{[\mathbf{Q}(\zeta):\mathbf{Q}]}{[\mathbf{Q}(\zeta^p):\mathbf{Q}]}=\frac{\varphi(n)}{\varphi(n/p)}=p,$$
so that every coefficient in this linear combination must be zero. (!!!) Hence for each $t$ with $0\le t\le p-1$, the sum of the values $a_i (\zeta^p)^{d_i}$ over all $i$'s with $e_i=t$ must be zero. And so on. The whole proof takes less than a page.
Best Answer
Using the Newton identities, one can (in the high characteristic regime $p>k$) express the elementary symmetric polynomial $\sum_{i_1 < \dots < i_k} a_{i_1} \dots a_{i_k}$ in terms of the moments $\sum_{i=1}^k a_i^j$ for $j=1,\dots,k$. The question then boils down to the equidistribution of $\sum_{i=1}^k (a_i^j)_{j=1}^k$ in ${\mathbf F}_p^k$. There is however an obstruction to equidistribution because the points $(a^j)_{j=1}^k$ for $a=-2,-1,1,2$ do not span the entirety of ${\mathbf F}_p^k$ once $k \geq 4$, due to the existence of non-trivial polynomials $P$ of degree $4$ or higher that vanish at $-2,-1,1,2$. For instance because $P(x) = (x+2)(x+1)(x-1)(x-2) = x^4 - 5 x^2 + 4$ vanishes at these points, the vector $(m_j)_{j=1}^k := \sum_{i=1}^k (a_i^j)_{j=1}^k$ is surely constrained to the hyperplane $m_4 - 5 m_2 + 4k = 0$. This is going to create some distortions to the probability that $\sum_{i_1 < \dots < i_k} a_{i_1} \dots a_{i_k}$ vanishes mod $p$ that can be explicitly calculated for each $p,k$, but the formula is going to be messy (these errors will be of lower order in the limit $p \to \infty$ holding $k$ fixed though, by the Lang-Weil estimates).
(To put it another way, one can use the Newton identities to express the elementary symmetric polynomial as some polynoimal combination of the frequencies $d_{-2}, d_{-1}, d_1, d_2$ that count how often the random sequence $a_1,\dots,a_k$ attains each of its permitted values $-2, -1, 1, 2$. To count the probability that this polynomial vanishes mod $p$, one has to count the points in some variety over ${\bf F}_p$ which in general is a task for the Lang-Weil estimate.)
Your second sum seems to be expressible in terms of a symmetric polynomial in a suitable matrix algebra. If one had $i_k < j_k$ in place of the constraint $i_k \leq j_k$ then one just needs to take the order $2k$ elementary symmetric polynomial of the matrices $\begin{pmatrix} 0 & b_i \\ a_i & 0 \end{pmatrix}$ and extract the top left coefficient. With $i_k \leq j_k$ the situation is more complicated but I expect there is still some sort of matrix representation. However being non-abelian I doubt there is a reduction to the abelian elementary symmetric polynomial considered earlier, and given how complicated that formula already was I'm afraid it is not going to be fun to try to control this sum (except possibly in the asymptotic regime where $p$ is somewhat large and $k$ goes to infinity; actually the nonabelian case might be substantially more "mixing" than the abelian one and one could conceivably get better asymptotics by using some of the theory of expansion of Cayley graphs, but this looks like a lot of work...).