Let $|A| = m \ge 4n\sqrt{n}$ and let $S$ be the set of all segments with endpoints in $A$. Clearly, $|S| = \binom{m}{2}$. The coordinates of every midpoint of a segment from $S$ are integer multiples of $1/2$. In the convex hull of $G$ there are less than $4n^2$ such points so there exists a point $B$ which is a midpoint of at least $(m/2)/(4n^2)$ segments from $S$. Let $P$ be the set of all segments from $S$ with their midpoints in $B$. Then
$$|P| \ge \frac{\binom{m}{2}}{4n^2} = \frac{m(m-1)}{8n^2} \ge \frac{4n\sqrt{n}(4n\sqrt{n}-1)}{8n^2} = \frac{16n^3 - 4n\sqrt{n}}{8n^2} = 2n - \frac{1}{2\sqrt{n}} > 2n-1.$$
and $|P| \ge 2n$.

Let us divide $P$ into disjoint families of segments which lay on the same line. Suppose that the number of such families is $k$ and in the $i$-th family we have $a_i$ segments, for $i = 1, \dots, k$. Every segment from $a_i$ segments of one family has its endpoints in $G$ and they have a common midpoint, so $a_i \le n/2$. Moreover every two segments from $P$ are the diagonals of a parallelogram iff they do not lay on the same line. Therefore for the number of different parallelograms with diagonals belonging to $P$ we have
$$\sum_{1 \le i < j \le k} a_i a_j = \frac{1}{2} \left( \left( \sum_{i=1}^{k} a_i \right)^2 - \sum_{i=1}^{k} a_i^2 \right) \ge \frac{1}{2} \left( \left( \sum_{i=1}^{k} a_i \right)^2 - \sum_{i=1}^{k} a_i \cdot \frac{n}{2} \right) = \frac{1}{2} \left( |P|^2 - |P| \cdot \frac{n}{2} \right) = \frac{1}{2} |P| \left( |P| - \frac{n}{2} \right) \ge n \left( 2n - \frac{n}{2} \right) = \frac{3}{2} n^2 > n^2.$$
Thus we have more than $n^2$ convex quadrangles (parallelograms) satisfying the given condition.