Double asymptotics: $E\|\mathbf{x}_n-\mathbf{z}\|_{\infty}^2 \to 0 \implies p^{-1}\mathbf{x}_n^\top\mathbf{x}_n\to 1$ as $(p,n) \to (\infty, \infty)$

Question

This is an advanced question for this.

Let $\mathbf{x}$ be a $p$-dimensional vector, and $\mathbf{z}$ is a $p$-dimensional standard normal vector whose elements are i.i.d. Now we assume for some scaling parameter $n \to \infty$ (and fixed $p$), $$ \mathbb{E} \|\mathbf{x} - \mathbf{z}\|_{\infty}^2 \to 0, $$ which is just says $\mathbb{E} [\max_{i\le p} |x_i - z_i|^2] \to 0$. Then my question is that the following.

Question.

Can we say that under this condition, for any path $(p,n) \to (\infty, \infty)$, $$ \frac{1}{p} \mathbf{x}^\top \mathbf{x} \xrightarrow{\mathbb{P}} 1?$$

For one path: $n\to \infty$ first, then $p \to \infty$ is quite obvious, but I'm not sure how to handle for any arbitrary path.

I look forward your any help.

Thanks,