We use the famous Frobenius endomorphism. Let $n = \sum_{i = 0}^k 2^{a_i}$ with $a_0 < a_1 < \cdots < a_k$ be the binary representation of $n$, and notice that\[(x^2 - x + 1)^{2^a}\equiv (x^2 + x + 1)^{2^a}\equiv x^{2^{a + 1}} + x^{2^a} + 1\pmod{2}\]by iterated Frobenius. Thus our polynomial becomes\[\prod_{i = 0}^k (x^{2^{a_i + 1}} + x^{2^{a_i}} + 1)\pmod{2}.\]Now, for any $m$ we notice that the coefficient of $x^m$ in this polynomial is the reduction modulo two of the number of ways to write $m = \sum_{i = 0}^k c_i2^{a_i}$, where each $c_i\in\{0, 1, 2\}$. We fix an $m$ and count such representations modulo two. For convenience, denote $a_{k + 1} = \infty$. We build it up from the bottom. At any point, after $c_0, \ldots, c_{j - 1}$ are fixed, consider $m - \sum_{i = 0}^{j - 1} c_i2^{a_i} \pmod{2^{a_j + 1}}$. This will uniquely determine $c_j \pmod{2}$, and if it is odd then clearly $c_j = 1$ is forced. Otherwise, $c_j\in\{0, 2\}$ and we have two cases. Case 1: $a_{j + 1} > a_j + 1$. Then taking the above quantity but $\pmod{2^{a_j + 2}}$ will fix $c_j \pmod{4}$, and then the value of $c_j$ is forced (or might not exist!). Case 2: $a_{j + 1} = a_j + 1$. Then consider all six possibilities $(c_j, c_{j + 1}) = (0, 0), (2, 2), (0, 1), (2, 0), (0, 2), (2, 1)$. Notice that $(0, 1), (2, 0)$ pair up, as do $(0, 2), (2, 1)$; that is, choosing $(0, 1)$ contributes a sum of $2^{a_j + 1}$ while so does $(2, 0)$. A similar phenomenon happens with $(0, 2), (2, 1)$. Thus, any valid representations we get out of choosing $(0, 1)$ are precisely mirrored by choosing $(2, 0)$, and the same for $(0, 2), (2, 1)$. Since we are counting modulo two, we can thus disregard all of those cases. Now $(c_j, c_{j + 1}) = (0, 0), (2, 2)$. These two cases contribute $2^{a_j}, 2^{a_j} + 2^{a_{j + 1}} = 3\cdot 2^{a_j}$ respectively to our sum, and finally again $\pmod{2^{a_j + 2}}$ will distinguish the unique valid case. We can follow this process for any value of $m$ and check the end result. We will either build up a unique representation of $m = \sum_{i = 0}^k c_i2^{a_i}$ by following the rules above, or we will fail at some point. Thus, $m$ has an odd number of overall representations (and therefore odd coefficient of $x^m$) if and only if it has such a unique representation according to our rules above. And now we simply count how many such numbers exist by building all of them simultaneously; notice that the above rules make it clear that only the sizes of each block of $1$s in the binary representation of $n$ matter, not their orders, so let them be $b_1, b_2, b_3, \ldots, b_l$. Then the number of ways to "build" a number $m$ with such a representation out of the block of length $b_i$ is $\frac{2^{b_i + 2} - (-1)^{b_i}}{3}$, as verified by simple induction (essentially, a block of length $b$ has $x_b = x_{b - 1} + 2x_{b - 2}$ way to do this and $x_1 = 3, x_2 = 5$). Indeed, a block $111$ gives us $11$ ways to assign the values $c_i$:\[(0, 0, 0), (1, 0, 0), (2, 0, 0), (0, 0, 1), (0, 1, 1), (1, 1, 1), (2, 1, 1), (2, 2, 1), (0, 2, 2), (1, 2, 2), (2, 2, 2)\] Multiplying gives\[\prod_{i = 1}^l\left(\frac{2^{b_i + 2} - (-1)^{b_i}}{3}\right),\]which is equivalent to what MellowMelon said.