$\mathbb Z^n/\langle (a,...,a) \rangle \cong \mathbb Z^{n-1} \oplus \mathbb Z/\langle a \rangle$

The following: $$(1,0,\dots,0), (0,1,0,\dots,0), \dots, (0,\dots,0,1,0), (1,1,\dots, 1)$$ is a basis of $\mathbb{Z}^n$ and $(a,\dots,a)$ corresponds to $(0,\dots,0,a)$ in this new basis. Your group is thus the isomorphic to the quotient: $$\mathbb{Z}^n / \langle (0, \dots, 0, a) \rangle \cong \mathbb{Z}^{n-1} \oplus \mathbb{Z}/(a).$$

If you want to use the first isomorphism theorem, then define $\psi:\mathbb Z^n \to \mathbb Z^{n-1} \oplus \mathbb Z/\langle a \rangle$ by $\psi(x_1,\dots,x_n)=(x_2-x_1,\dots,x_n-x_1,\bar x_1)$. Check that $\psi$ is a surjective homomorphism and $\ker\psi=\langle (a,\dots,a) \rangle$.

$\mathbb{Z}^n/\langle (a,\dotsc,a)$ is an abelian group with the presentation $\langle e_1,\dotsc,e_n : a e_1 + \cdots + a e_n = 0\}$. Observe that $e_1,\dotsc,e_{n-1},e_1+\cdots+e_n$ is also a generating system and $a(e_1+\cdots+e_n)=0$. Hence, we obtain a homomorphism from $\langle f_1,\dotsc,f_n : a f_n = 0 \rangle$. One then constructs its inverse.