If $x = (x_1,x_2, \dots, x_n) \in \mathbb{R}^n$, and
$$ x^Tx = \begin{bmatrix} x_1^2 & x_1x_2 & x_1x_3 & \dots & x_1x_n \\ x_2x_1 & x_2^2 & x_2x_3 & \dots & x_2x_n \\ \vdots & & \vdots & & \vdots \\ x_nx_1 & x_n x_2 & \dots & \dots & x_n^2\end{bmatrix}, $$
is it possible to show that $||x^Tx||$ is comparable to $|x|^2 = tr(x^Tx)$?
According to the notation I like to use ($x$ is a column vector and the matrix multiplication is on the left of the vector), the big matrix you wrote is $xx^T$ (while $x^Tx$ is the scalar $\lvert x\rvert^2$). I'll stick to mine because I don't want ot mess up, but it's basically the same thing up to transposition. If $x\ne0$, the matrix $xx^T$ is orthogonally diagonalisable. Per se, this is a consequence of spectral theorem, but for the purpose of the question we need to prove it manually anyways. In fact, $\ker xx^T=x^{\perp}$. Moreover $$(xx^T)x=x(x^Tx)=\lvert x\rvert^2x$$
So, if $$P=\left(x,b_1,\cdots,b_{n-1}\right)$$ where $b_1,\cdots,b_{n-1}$ is an orthonormal basis of $\ker xx^T=x^{\perp}$, the matrix $P$ is orthogonal and $P^{-1}xx^TP=\operatorname{diag}(\lvert x\rvert^2,0,\cdots,0)$. Since the operator norm is invariant under conjugation by an orthogonal matrix, you have that $$\lVert xx^T\rVert=\lvert x\rvert^2=\operatorname{tr}(xx^T)$$