Essentially the centers of the circles form an equilateral triangle of sidelength $r$. This is a pretty well known problem, the answer is $3(\frac{\pi}{6}r^2-\frac{\sqrt{3}}{4}r^2)+\frac{\sqrt{3}}{4}r^2$

Suppose that the three centers are $A,$ $B,$ and $C.$ Let's focus on the desired region. We see that when we take the sum $$\text{[Sector ABC]}+\text{[Sector BCA]}+\text{[Sector CAB]}=3\text{[Sector ABC]},$$every part of our desired region is counted exactly once except for the central equilateral triangle (Namely $\triangle ABC$), which is counted three times. Therefore, the area of our desired region is equal to $$3\text{[Sector ABC]}-2[ABC].$$Sector $ABC$ has a central angle of $60^\circ,$ and so $$3\text{[Sector ABC]} = 3 \cdot \frac{n^2 \pi}{6} = \frac{n^2 \pi}{2}.$$$\triangle ABC$ has a sidelength of $n,$ and so $$2[ABC]=2 \cdot \frac{n^2 \sqrt{3}}{4} = \frac{n^2 \sqrt{3}}{2}.$$Therefore, the shaded region is equal to $$\frac{n^2 \pi}{2} - \frac{n^2 \sqrt{3}}{2} = n^2\left(\frac{\pi-\sqrt{3}}{2}\right).$$