Note that $ \lfloor \tau^0 \rceil = \lfloor 1 \rceil = 1 $ and $ \lfloor \tau \rceil = 2 $. Let $ \sigma = \frac{1 - \sqrt{5}}{2}, $ and let $ L_n = \tau^n + \sigma^n $. Then \begin{align*} L_n &= \tau^n + \sigma^n \\ &= (\tau + \sigma)(\tau^{n - 1} + \sigma^{n - 1}) - \tau \sigma (\tau^{n - 2} + \sigma^{n - 2}) \\ &= L_{n - 1} + L_{n - 2}.\end{align*}Also, $ L_0 = 2 $ and $ L_2 = 1, $ so $ L_n $ is an integer for all $ n \ge 0 $. Furthermore, \[\sigma^2 = \frac{3 - \sqrt{5}}{2} < \frac{1}{2},\]so for $ n \ge 2, $ $ |\sigma^n| < \frac{1}{2} $. Hence, \[\lfloor \tau^n \rceil = L_n\]for all $ n \ge 2 $. Let \[S = \frac{L_2}{2^2} + \frac{L_3}{2^3} + \frac{L_4}{2^4} + \dotsb.\]Then \begin{align*} S &= \frac{L_2}{2^2} + \frac{L_3}{2^3} + \frac{L_4}{2^4} + \dotsb \\ &= \frac{L_0 + L_1}{2^2} + \frac{L_1 + L_2}{2^3} + \frac{L_2 + L_3}{2^4} + \dotsb \\ &= \left( \frac{L_0}{2^2} + \frac{L_1}{2^3} + \frac{L_2}{2^4} + \dotsb \right) + \left( \frac{L_1}{2^2} + \frac{L_2}{2^3} + \frac{L_3}{2^4} + \dotsb \right) \\ &=\left( \frac{1}{2} + \frac{1}{8} + \frac{S}{4} \right) + \left( \frac{1}{4} + \frac{S}{2} \right).\end{align*}Solving, we find $ S = \frac{7}{2} $. Therefore, \[\sum_{n = 0}^\infty \frac{\lfloor \tau^n \rceil}{2^n} = 1 + \frac{2}{2} + \frac{7}{2} = \boxed{\frac{11}{2}}.\]