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In the diagram above, notice that triangle $OAB$ and triangle $ABC$ are congruent and equilateral with side length $2$. We can see the radius of the larger circle is $2\overline{OH}+1$. Using $30^{\circ }-60^{\circ }-90^{\circ }$ triangles, we know $\overline{OH}=\sqrt{3}$. Therefore, the radius of the larger circle is $2\sqrt{3}+1$. The area of the larger circle is thus ${\left(2\sqrt{3}+1\right)}^2\pi =\left(13+4\sqrt{3}\right)\pi$, and the sum of the areas of the smaller circles is $13\pi$, so the area of the dark region is $\left(13+4\sqrt{3}\right)\pi -13\pi =\boxed{4\pi \sqrt{3}}$.