suppose that $\mathbf p = [p_1, p_2, ..., p_n]'$ is a random vector. (' == transpose) and each element of $\mathbf p$ like $p_i$ is a Gaussian random variable with zero mean ($\mathbb E(p_i)=0$) and variance $v_i$,($\mathbb E(p_i^2)=v_i$).
the elements of $\mathbf p$ are independent.
then what is the value of $\frac{(\mathbf p'\times \mathbf p)}{n}$ when $n$ goes to infinity?
I know that when elements of p are having the same variance, the answer is equal to the variance, but what is the answer when the variance of each element is different?
Notice that $$\frac{\boldsymbol p^{\top}\boldsymbol p}{n}=\frac{p_1^2+\cdots+p_n^2}{n}.$$ Given your comment about the variance, it seems to me that what you're really interested in is \begin{align}\tag{1} \text{E}\left[\frac{\boldsymbol p^{\top}\boldsymbol p}{n}\right]=\frac{v_i+\cdots+v_n}{n}. \end{align} Indeed, if $v_1=v_2=\cdots=v_n=v$, then it is true that $$\text{E}\left[\frac{\boldsymbol p^{\top}\boldsymbol p}{n}\right]=\frac{nv}{n}=v.$$
If the $v_i$ are not all equal to each other, there is unfortunately no single solution to $(1)$. However, there exists criterions that can help you find upper/lower bounds for the limit, and sometimes even find what the limit is.
A limit of the form $$\lim_{n\to\infty}\frac{\sum_{i=1}^na_i}n$$ where $\{a_n:n\in\mathbb N\}$ is a sequence in $\mathbb R$ is known as a Cesàro average (or Cesàro mean). One can show that if $a_n\to a$, then $\sum_ia_i/n\to a$ as well. I also recall seeing somewhere that, in general, $$\liminf_{n\to\infty}a_n\leq\liminf_{n\to\infty}\frac{\sum_{i=1}^na_i}n\leq\limsup_{n\to\infty}\frac{\sum_{i=1}^na_i}n\leq\limsup_{n\to\infty}a_n,$$ which may at least give bounds for $(1)$ (I think this one is called the Stolz-Cesàro theorem).