Let $A\in\mathbb{R}^{n^k}$ be a $k$-dimensional tensor with $n$ elements along each dimension. Moreover suppose $u_1,u_2,\dots,u_k\sim\text{Unif}(\pm1)^n$ are $n$ dimensional vectors with each of their components drawn iid from Rademacher distribution, and let $U:=u_1\otimes u_2 \otimes \dots \otimes u_k$ be a rank-1 random tensor constructed from these vectors. Finally, let $X:=\langle A, U \rangle$, in which by $\langle, \rangle$ we simply mean the inner product of tensor if we flatten them as vectors.

**Question:**: What is the concentration of $X^2$ around its mean? As it is shown at the bottom of the post, $\mathbb{E} X^2=\lVert A\rVert_F^2$, i.e., the sum of the square of all elements in the tensor. How can the tails of the distribution around the mean be bounded: $P\left(|X^2-\mathbb{E} X^2|\ge x \right)$? Alternatively, as is typical in many concentration bounds, can the probability be bounded as an $\epsilon$-deviation from the mean: $P\left(|X^2-\mathbb{E} X^2|\ge \epsilon \left(\mathbb{E} X^2\right) \right)$?

**Follow-up** as @IosifPinelis has pointed out in their answer, moments of $X$, $\mathbb{E} X^{2p}$, can be effectively bounded, leading to a tail bound roughly of type $P(X\ge x)\le \exp(-x^{2/k})$. However, the main question was to bound the tail bounds of $X^2$ around its mean, which is $P(|X^2-\mathbb{E} X^2| \ge x)$. Taking a similar approach as @IosifPinelis, one could try to bound the tails of $Y:=X^2 - \lVert A\rVert_F^2$ by bounding its moments $\mathbb{E} Y^p$. Is that a sound approach?

**Context:** given that tails are light, this can be seen as sketching the Frobenius norm of a tensor. More so to the point, if $A,B$ are tensors, if we compute $v=\langle A,u_1\otimes \dots\otimes u_k\rangle$ and $w=\langle B,u_1\otimes \dots\otimes u_k\rangle$, then $\mathbb{E}(v-w)^2 = \lVert A-B\rVert_F^2$ and if the concentration is high, it can be seen as a distance-preserving embedding of tensors into the $L^2$-norm. The reason for choosing a rank-1 random tensor as opposed to a fully random one, which makes the bound trivial to compute, is that for certain applications the product with a rank-1 tensor can be computed much more efficiently and hence the sketching is faster to compute.

**Derivation of $\mathbb{E} X^2$:**
\begin{align*}
\mathbb{E} X^2 &= \mathbb{E}\sum_{i_1,j_1, \dots,i_k,j_k} a_{i_1, \dots,i_k} a_{j_1,\dots,j_k} u_{1,i_1}\dots u_{k,i_k} u_{1,j_1}\dots u_{k,j_k}\\
&= \sum_{i_1,j_1, \dots,i_k,j_k\le n} a_{i_1, \dots,i_k} a_{j_1,\dots,j_k} \Pi_{m=1}^k \mathbb{E}(u_{m,i_m} u_{m,j_m}) &&\rhd \text{L.o.E, and $u_i$s independence}\\
&= \sum_{i_1,\dots,i_k\le n} a_{i_1,\dots,i_k}^2 && \rhd i_m\neq j_m: \mathbb{E} u_{i_m}u_{j_m} = 0\\
&= \sum_{i_1,\dots,i_k\le n} a_{i_1,\dots,i_k}^2=\lVert A \rVert_F^2 && \rhd \mathbb{E} u_{i_m}^2 = \mathrm{Var}(u_{i_m}) = 1
\end{align*}