Dear Daniel, here is a detailed explanation, respectfully following the sacred texts (EGA or Hartshorne).

a) First of all, $\mathbb A^{n+1}_{\mathbb Z}$ has no origin, despite our classical intuition! As a substitute, it has the prime (but not maximal) ideal $\mathcal P=(X_1,X_2,...,X_{n+1})$ and corresponding to it the integral subscheme $V=V(\mathcal P)$. And what you want to calculate is the set of $\mathbb Z$-points of
$U=\mathbb A^{n+1}\setminus V(\mathcal P)$, i.e. the set of morphisms $Spec(\mathbb Z)\to U$. Let's do that.

b) A morphisms $f: Spec \mathbb Z \to \mathbb A^{n+1}$ corresponds to a morphism of rings
$ev_a: \mathbb Z[X_1,X_2,...,X_{n+1}] \to \mathbb Z$ , evaluation of integral polynomials at a tuple $a=(a_1,a_2,...,a_{n+1}) \in {\mathbb Z}^{n+1} $. Call $f=f_a : Spec \mathbb Z \to \mathbb A^{n+1}$ the corresponding morphism.

c) We must ensure that the image of $f$ lies in $U$ i.e. that it is disjoint from
$V=V(\mathcal P)$. But the points of $V$ are its generic point $(X_1,X_2,...,X_{n+1})$ and its closed points $\mathcal M_p=(X_1,X_2,...,X_{n+1}, p)$, $p$ a prime. It is enough to show that these closed points are not in the image of $f$. Equivalently, we must show that the fibre of $f$ at $\mathcal M_p$ is empty. Since this fibre is the spectrum of
$\mathbb Z / (a_1,a_2,...,a_{n+1},p)$ our condition is that this ring be zero or equivalently that the ideal $(a_1,a_2,...,a_{n+1},p)$ be zero: this will happen exactly if $p$ does not divide all ot the $a_i$'s. Since this must hold for all primes, we get:

**Final result** The $\mathbb Z$-points of $U=\mathbb A^{n+1}\setminus V(\mathcal P)$ are given by $(n+1)$-tuples of integers whose g.c.d. is 1.

**Reminder** I have used that the fibre of a morphism of affine schemes $f:SpecB\to Spec A$ at $\mathcal M \in Specmax A$ is $Spec ( B/\mathcal M B) $.