The short answer : The geometric points of the projective line $\mathbb{P}^1$ over the ‘field with one element’ $\mathbb{F}_1$ form the set $\{ 0,\infty \} \cup \pmb{\mu}$ with $\pmb{\mu}$ the group of all roots of unity. Its schematic points form the set $\{ \infty,0 \} \cup \{ [1],[2],[3],[4],\cdots \}$ and the degree of the point $[n]$ equals $\phi(n)$.
The longer answer : We have seen that the geometric points of $\mathbb{P}^1$ over the finite field $\mathbb{F}_p$ form the set
$$\{ 0 = [0:1],~\infty=[1:0] \} \cup \{ \alpha = [\alpha:1]~:~\alpha \in \overline{\mathbb{F}}_p^{\ast} \}$$
Claim : the multiplicative group of the non-zero elements of the algebraic closure $\overline{\mathbb{F}}_p^{\ast}$ is isomorphic as group to the group $\pmb{\mu}^{(p)}$ of all roots of unity of order prime to $p$.
Clearly, any element $\alpha \in \mathbb{F}^{\ast}_{p^n}$ has order some divisor of $p^n-1$ and hence is prime to $p$. Conversely, if $(m,p)=1$ then $\overline{p}$ is a unit in $\mathbb{Z}/m\mathbb{Z}$ and therefore for some $n$ we have $p^n \cong 1~mod(m)$. But then, $m | p^n-1$ and there are primitive $m$-th roots of unity in $\mathbb{F}_{p^n}$.
However, describing this correspondence explicitly from a given construction of $\overline{\mathbb{F}}_p$ is very challenging. For example, John Conway proved in ONAG that $\overline{\mathbb{F}}_2$ can be identified with all ordinals smaller than $\omega^{\omega^{\omega}}$ equipped with nim-addition and multiplication.
Finding the correspondence between small ordinals and odd roots of unity is the topic of the post The odd knights of the round table (and follow-up posts here and here).
Below is the correpondence between $\mathbb{F}_{2^4}^{\ast}$ (identified with the ordinals from $1$ to $15$) and the 15-th roots of unity (nim-addition and nim-multiplication tables on the left). The lines describe the involution $x \mapsto x+1$.
The schematic points of $\mathbb{P}^1$ over $\mathbb{F}_p$ is the set of all $Gal(\overline{\mathbb{F}}_p/\mathbb{F}_p) = \hat{\mathbb{Z}}$-orbits on the geometric points, and the degree of a scheme-point is the number of geometric points in the orbit.
Assigning to such a Galois-orbit $\mathcal{O}$ the polynomial $\prod_{\alpha \in \mathcal{O}} (x-\alpha)$ identifies the schematic points of $\mathbb{P}^1/\mathbb{F}_p$ with all irreducible polynomials in $\mathbb{F}_p[x]$ (together with $\infty$) and the point-degree coincides with the degree of the polynomial.
Concrete : say we have an explicit identification of $\mathbb{F}_{p^n}^{\ast}$ with all $p^n-1$-th roots of unity, then we can find all irreducible polynomials in $\mathbb{F}_p[x]$ of degree a divisor of $n$ by studying the orbits of these roots of unity under the power-map $z \mapsto z^p$.
In the picture above, we have indicated the different orbits of $\mathbb{F}_{2^4}$ with different colors. There are two orbits of length one : $\{ 0 \}$ corresponding to $x$ and $\{ 1 \}$ corresponding to $x+1$. One orbit of length two $\{ 2,3 \}$ corresponding to the irreducible polynomial $x^2+x+1$ (check the tables to verfify that this is indeed $(x-2)(x-3)$) and three orbits of length four
$\{ 4,6,5,7 \} \leftrightarrow x^4+x+1$
$\{ 11,12,9,15 \} \leftrightarrow x^4+x^3+1$
$\{ 14,8,13,10 \} \leftrightarrow x^4+x^3+x^2+x+1$
By analogy we can now define the geometric points of $\mathbb{P}^1$ over the field with one element $\mathbb{F}_1$ to be the set $\{ 0,\infty \} \cup \pmb{\mu}^{(1)}$ where $\pmb{\mu}^{(1)}$ are all roots of unity of order prime to $1$, that is just all of them : $\pmb{\mu}$.
The schematic points of $\mathbb{P}^1/\mathbb{F}_1$ are then the orbits of this set under the action of the Galois group $Gal(\mathbb{Q}(\pmb{\mu})/\mathbb{Q})$.
One checks that these orbits correspond to $\{ 0,\infty \}$ and $\{ [1],[2],[3],[4],\cdots \}$ where $[n]$ is the orbit consisting of all primitive $n$-th roots of unity. Consequently, the degree of the scheme-point $[n]$ is equal to $\phi(n)$ with $\phi$ the Euler function.
