Suppose you have a unit square and want to completely cover it with $1$ circle ($n=1$). If you want to minimize the excess area, the area inside of the circle and outside of the square, you would create the circle such that each vertex of the square lands on the circle. The excess area is highlighted in red. The excess area is the area highlighted in red which is $\frac{\pi}{2}-1$.
For the case $n=2$, it is a little trickier. I came up with this, which I think minimizes excess area at $\frac{5}{8}\left(\pi-\tan^{-1}2-\frac{6}{5}\right)$.
What is the minimum excess area for $n$ circles?
The excess area is not necessarily equal to the sum of the excess area of each circle. It is equal to the excess area of the union of all the circles.