I am looking for an orthonormal matrix $A\in\mathbb{C}^{n\times n}$, such that for any two elements $a_{i\ j},a_{p\ q}$ ( not necessarily the distinct elements ) of $A$, we have that:
\begin{equation} |a_{i\ j}.\overline{a_{p\ q}}| < \frac{1}{n} \end{equation}
Can anybody tell me if such a matrix exists and how to find it?
First, your condition can be simplified: it's actually equivalent to requiring $|a_{jk}| < \frac{1}{\sqrt{n}}$ for all $j,k$. Proof: If all $a_{jk}$ satisfy $|a_{jk}| < \frac{1}{\sqrt n}$ then we'll clearly always have $|a_{jk} \bar{a_{pq}}| = |a_{jk}| |a_{pq}| < \left( \frac 1 {\sqrt{n}} \right)^2 = \frac 1 n$. In the other direction, if your version of the condition holds, then choosing $p=j$ and $q=k$ we find $|a_{jk} \bar{a_{jk}}| < \frac 1 n$, so $|a_{jk}| < \frac 1 {\sqrt n}$.
Second, the answer to your question is that no such matrix can exist. By definition of an orthonormal matrix, we'd need $\sum_k |a_{jk}|^2 = 1$ for each row $j$, so at least one of the entries in that row must have $|a_{jk}|^2 \ge \frac 1 n$ and thus $|a_{jk}| \ge \frac 1 {\sqrt n}$. That contradicts the condition I just proved is equivalent to yours. (Credit to @user1551 who wrote a short version of this argument in a comment.)
PS: When working with complex numbers, I'd really recommend using $j,k$ as your default indices instead of $i,j$! It's too easy to misread my own work when some $i$'s are integer indices and others are $\sqrt{-1}$.