Let's first factor $f(x) = x^{13} + 1$ into irreducible factors. It's clear that $-1$ is a zero, and so the linear factor $x + 1$ divides $x^{13} + 1$. Dividing by $x + 1$ gives $f(x) = x^{13} + 1 = (x + 1)(x^{12} - x^{11} + x^{10} - \dots + 1)$.
Now take \begin{align*} g(x) &= (x - 1)^{13} + 1 \\ &= \sum_{k = 0}^{13} \binom{13}{k} (-1)^{13 - k}x^k + 1 \\ &= x \sum_{k = 0}^{12} \binom{13}{k + 1} (-1)^{13 - (k + 1)}x^k \end{align*}
Now when $k = 12$, then $\binom{13}{k + 1} = 1$, and for $0 \leq k < 12$, $13 \mid \binom{13}{k + 1}$, but $13^2 \nmid \binom{13}{1}$, and so by Eisenstein's criterion, the polynomial represented by the sum is irreducible, and so we have the factorization $g = xp$, with $p$ irreducible. Now since $f(x) = g(x + 1)$, we know that $f$ factors into $f = (x + 1)q$, with $q$ irreducible, and as we have already seen $q$ must equal $(x^{12} - x^{11} + x^{10} - \dots + 1)$, which is thus irreducible.
Now take any zero of $\zeta$ of $q$, and consider the field $\mathbb{Q}(\zeta)$. Since $q$ is irreducible, the elements of the form $1,\zeta,\zeta^2,\dots,\zeta^{11}$ are linearly independent (for otherwise $\zeta$ would be a zero of a polynomial of smaller degree, contradicting the minimality of $q$), and since $\zeta^{12} = \zeta^{11} - \zeta^{10} + \dots - 1$, any higher power of $\zeta$ can be written as a linear combination in $1,\zeta,\zeta^2,\dots,\zeta^{11}$. This proves that $\mathbb{Q}(\zeta)$ has degree 12. Now for each $0 < k \leq 12$, take the elements $ \zeta_k = \begin{cases} \zeta^k & \text{when $k$ is odd} \\ -\zeta^k & \text{when $k$ is even} \end{cases} $ Clearly $\zeta_k^{13} = -1$, and these $\zeta_k$ must all be different (for otherwise, we must either have $\zeta_{12} = \zeta_0 = 1$, but we know that's not true, or otherwise, if $\zeta_i = \zeta_j$, with $i \neq j$, then $\zeta$ would be a zero of a polynomial of degree lower than 12, which contradicts the minimality of $q$). Since we now have 13 distinct zeros, we know that $\mathbb{Q}(\zeta)$ is the splitting field of $x^{13} + 1$, and as we have already seen the degree $[\mathbb{Q}(\zeta) : \mathbb{Q}]$ is 12.