If $ x $ is a term in the sequence, then the next term is $ x^3 - 3x^2 + 3 $. These are equal if and only if \[x^3 - 3x^2 + 3 = x,\]or $ x^3 - 3x^2 - x + 3 = 0 $. This factors as $ (x - 3)(x - 1)(x + 1) = 0, $ so $ x = 3, $ $ x = 1, $ or $ x = -1 $. Furthermore, using this factorization, we can show that if $ a_n > 3, $ then $ a_{n + 1} = a_n^3 - 3a_n^2 + 3 > a_n, $ and if $ a_n < -1, $ then $ a_{n + 1} = a_n^3 - 3a_n^2 + 3 < a_n, $ so any possible values of $ a_0 $ must lie in the interval $ [-1,3] $. Thus, we can let \[a_0 = 1 + 2 \cos \theta = 1 + e^{i \theta} + e^{-i \theta},\]where $ 0 \le \theta \le \pi $. Then \begin{align*} a_1 &= a_0^3 - 3a_0^2 + 3 \\ &= (a_0 - 1)^3 - 3a_0 + 4 \\ &= (e^{i \theta} + e^{-i \theta})^3 - 3(1 + e^{i \theta} + e^{- i\theta}) + 4 \\ &= e^{3i \theta} + 3e^{i \theta} + 3e^{-i \theta} + e^{-3i \theta} - 3 - 3e^{i \theta} - 3e^{-i \theta} + 4 \\ &= 1 + e^{3i \theta} + e^{-3i \theta}.\end{align*}In general, \[a_n = 1 + e^{3^n i \theta} + e^{-3^n i \theta}.\]In particular, $ a_{2007} = 1 + e^{3^{2007} i \theta} + e^{-3^{2007} i \theta} = 1 + 2 \cos 3^{2007} \theta $. We want this to equal $ 1 + 2 \cos \theta, $ so \[\cos 3^{2007} \theta = \cos \theta.\]In other words, \[\cos 3^{2007} \theta - \cos \theta = -2 \sin \frac{(3^{2007} + 1) \theta}{2} \sin \frac{(3^{2007} - 1) \theta}{2} = 0.\]If $ \sin \frac{(3^{2007} + 1) \theta}{2} = 0, $ then $ (3^{2007} + 1) \theta = 2n \pi $ for some integer $ n $. The possible values of $ n $ are 0, 1, $ \dots, $ $ \frac{3^{2007} + 1}{2}, $ giving us $ \frac{3^{2007} + 1}{2} + 1 $ solutions. If $ \sin \frac{(3^{2007} - 1) \theta}{2} = 0, $ then $ (3^{2007} - 1) \theta = 2n \pi $ for some integer $ n $. The possible values of $ n $ are 0, 1, $ \dots, $ $ \frac{3^{2007} - 1}{2}, $ giving us $ \frac{3^{2007} - 1}{2} + 1 $ solutions. The two family of solutions include 0 and $ \pi $ twice, so the total number of solutions is \[\frac{3^{2007} + 1}{2} + 1 + \frac{3^{2007} - 1}{2} + 1 - 2 = \boxed{3^{2007}}.\]