Setting $ z = re^{i \theta} $ in the given equation, we get \[\left| 2re^{i \theta} + \frac{1}{r} e^{-i \theta} \right| = 1.\]Then \[\left| 2r \cos \theta + 2ri \sin \theta + \frac{1}{r} \cos \theta - \frac{i}{r} \sin \theta \right| = 1.\]Thus, \[\left( 2r \cos \theta + \frac{1}{r} \cos \theta \right)^2 + \left( 2r \sin \theta - \frac{1}{r} \sin \theta \right)^2 = 1.\]Expanding, we get \[4r^2 \cos^2 \theta + 4 \cos^2 \theta + \frac{1}{r^2} \cos^2 \theta + 4r^2 \sin^2 \theta - 4 \sin^2 \theta + \frac{1}{r^2} \sin^2 \theta = 1,\]which simplifies to \[4r^2 + 4 \cos^2 \theta - 4 \sin^2 \theta + \frac{1}{r^2} = 1.\]Since $ \cos^2 \theta = 1 - \sin^2 \theta, $ \[4r^2 + 4 - 4 \sin^2 \theta - 4 \sin^2 \theta + \frac{1}{r^2} = 1,\]so \[8 \sin^2 \theta = 4r^2 + \frac{1}{r^2} + 3.\]By AM-GM, $ 4r^2 + \frac{1}{r^2} \ge 2 \sqrt{4r^2 \cdot \frac{1}{r^2}} = 4, $ so $ 8 \sin^2 \ge 7, $ or \[\sin^2 \theta \ge \frac{7}{8}.\]Equality occurs when $ r = \frac{1}{\sqrt{2}}, $ so the minimum value of $ \sin^2 \theta $ is $ \boxed{\frac{7}{8}} $.