Here are three ways to prove that the degree of $\zeta_{12}$ over $\Bbb{Q}$ is not 2. Two of them will actually prove to you that the degree of this over the rationals is actually 4.
$\textbf{Approach 1:}$ Let's do this in a very simple way and try to determine the degree of $\zeta_{12}$ over $\Bbb{Q}$. Notice that
$$\Bbb{Q}(\zeta_{12}) \cong \Bbb{Q}(\sqrt{3},i) \cong \left(\Bbb{Q}(\sqrt{3})\right)(i). $$
Now the degree of $i$ over $\Bbb{Q}$ is 2 because the minimal polynomial of $i$ over $\Bbb{Q}$ is $x^2 +1$. Now we show that the degree of $\sqrt{3}$ over $\Bbb{Q}(i)$ is two as well. We already have a polynomial with coefficients in $\Bbb{Q}(i)$ with $\sqrt{3}$ as a root, namely the polynomial $x^2 - 3$. Now suppose that this polynomial is reducible, meaning that $\sqrt{3} \in \Bbb{Q}(i)$. Then we get that since as a vector space over $\Bbb{Q}$, $\Bbb{Q}(i)$ is two dimensional we can write
$$\sqrt{3} = a + bi$$
for some $a,b \in \Bbb{Q}$. Then squaring both sides we get that $3 = a^2 - b^2 + 2abi$ and hence that $$\frac{3 - a^2 + b^2}{2ab} = i$$
in particular that $i$ is real. This is a contradiction so that the polynomial $x^2 - 3$ is irreducible, and hence that $[\Bbb{Q}(\sqrt{3},i) : \Bbb{Q}(i)] = 2$. It follows by the dimension counting formula that $[\Bbb{Q}(\sqrt{3},i):\Bbb{Q}] = 2 \times 2 = 4$ and hence the degree of the minimal polynomial of $\zeta_{12}$ over $\Bbb{Q}$ must be 4 and you are right. Therefore if you have found a monic degree polynomial of degree 4 with $\zeta_{12}$ as a root, it must be the case that that polynomial is irreducible otherwise this would contradict our result that $[\Bbb{Q}(\zeta_{12}):\Bbb{Q}] = 4$.
$\textbf{Add-on to approach 1:}$ Suppose you did not know about cyclotomic polynomials (like me) and wanted to compute the minimal polynomial of $\zeta_12$ over $\Bbb{Q}$. You already know that it must be of degree 4 by the argument above. Once you find a monic one of degree 4, it must be unique by uniqueness of the minimal polynomial (exercise). Write
$$x = \frac{\sqrt{3} + i}{2}.$$
Then $2x = \sqrt{3} + i$ so that squaring both sides gives that $4x^2 = 3 - 1 + 2\sqrt{i}$ which implies that $4x^2 - 2 = 2\sqrt{3}i$. Hence
$$\begin{eqnarray*} 2x^2 - 1 &=& \sqrt{3}i\\
\implies 4x^4 - 4x^2 + 1 &=& -3 \\
\implies 4x^4 - 4x^2 + 4 &=& 0 \\
\implies x^4 - x^2 +1 &=& 0 \\
\end{eqnarray*}$$
and voilà! This is exactly the cyclotomic polynomial of degree 4 that you found.
$\textbf{Approach 2: Galois Theory}$ Suppose that $[\Bbb{Q}(\zeta_{12}) : \Bbb{Q}] = 2$. Then this is a degree 2 extension of a field of characteristic zero and hence is a Galois Extension. It follows that the Galois Group $ \operatorname{Gal}(\Bbb{Q}(\zeta_{12})/\Bbb{Q}) \cong C_2$, the cyclic group of order 2.
Now take some $\sigma \in G$. Then $\sigma$ must induce a permutation on the roots of the minimal polynomial of $\zeta_{12}$ over $\Bbb{Q}$. But then since $\sigma(\sqrt{3})^2 = \sigma(3) = 3$ which implies that $\sigma$ fixes $\sqrt{3}$ (similarly it fixes $i$) it follows that
$$\sigma(\zeta_{12}) = \zeta_{12}.$$
But then since $\sigma \in \operatorname{Gal}(\Bbb{Q}(\zeta_{12})/\Bbb{Q})$ was arbitrary this means that every $ \sigma \in \operatorname{Gal}(\Bbb{Q}(\zeta_{12})/\Bbb{Q})$ is the identity permutation so that $ \operatorname{Gal}(\Bbb{Q}(\zeta_{12})/\Bbb{Q})$ is the trivial group. But then this contradicts the fact that $ \operatorname{Gal}(\Bbb{Q}(\zeta_{12})/\Bbb{Q}) \cong C_2$ so it is not possible for $[\Bbb{Q}(\zeta_{12}):\Bbb{Q}]$ to be equal to 2.
$\textbf{Edit:}$ The proof above on Galois Theory is incorrect. Namely because it is not true that $\sigma(\sqrt{3}) = \sqrt{3}$ alone, but rather $\sigma(\sqrt{3}) = \pm \sqrt{3}$ and similarly for $\sigma(i)$.
$\textbf{Approach 3:}$ There is this result that is useful:
$$[\Bbb{Q}(\zeta_n) : \Bbb{Q}] = \varphi(n)$$
where $\varphi(n)$ is the Euler totient function. It counts the number of integers $k$ such that $1 \leq k \leq n$ with $k$ relatively prime to $n$. In your case, the integers relatively prime to 12 are $1,5,7,11$ which is $4$, exactly what you need.
Best Answer
Galois theory provides some machinery for this:
Suppose $\rm L/K$ is Galois with $\rm G=Gal(L/K)$ and $\rm m(x):=minpoly_{\alpha,K}(x)$. Then
$\quad \rm m(\sigma(\alpha))=\sigma(m(\alpha))=\sigma(0)=0$ implies $\rm (x-\sigma\alpha)\mid m$ in $\rm L[x]$ for all $\rm \sigma\in G$,
$\quad \rm(x-\sigma\alpha)$ all coprime, $\rm \sigma\in G/S$, $\rm S=Stab_G(\alpha)$, implies $\rm f(x):=\prod\limits_{\sigma\in G/S}(x-\sigma\alpha)\mid m$ in $\rm L[x]$,
$\quad \rm \sigma f(x)=f(x)$ for all $\sigma\in G$ implies $\rm f(x)\in K[x]$; $\rm f(\alpha)=0$ implies $\rm m(x)\mid f(x)$ in $\rm K[x]$,
$\quad \rm f(x)\mid m(x)$ and $\rm m(x)\mid f(x)$ and both $\rm f,m$ monic implies $\rm f(x)=m(x)$.
Therefore the zeros of $\rm\alpha$'s minimal polynomial over $\rm K$ are precisely its $\rm Gal(L/K)$-conjugates.
As ${\rm Gal}\big({\bf Q}(\zeta_n)/{\bf Q}\big)=({\bf Z}/n{\bf Z})^\times$ it suffices to consider $\sigma:\zeta\mapsto\zeta^k$ for $k=1,\cdots,12$.
By symmetry we need only consider $1,\cdots,6$ for $\alpha=\zeta+\zeta^{-1}$. Thus
$$\begin{array}{ll} {\rm minpoly}_{\zeta+\zeta^{-1},\bf Q}(x) = & ~~~~\left(x-(\zeta+\zeta^{-1})\right)\left(x-(\zeta^2+\zeta^{-2})\right)\left(x-(\zeta^3+\zeta^{-3})\right) \\ & \times\left(x-(\zeta^4+\zeta^{-4})\right)\left(x-(\zeta^5+\zeta^{-5})\right)\left(x-(\zeta^6+\zeta^{-6})\right). \end{array}$$
Simplify the resulting expansion via the rules $\zeta^n=\zeta^{n\,\bmod\,13}$ and $\sum\limits_{k=0}^{12}\zeta^k=0$.