Suppose we have an anti-circulant matrix $A = (a_{ik})_{n\times n}$, the first row of $A$ is $r_1 = [a_1, a_2, \cdots, a_n]$ and the $i$-th row is $r_i = [a_i, a_{i+1}, \cdots, a_n, a_1, \cdots, a_{i - 1}]$.
For this answer, I will first give a construction of eigenvectors and eigenvalues, then prove that all eigenvalues and eigenvectors can be obtained in this way.
For the first part, let's consider the DFT matrix $D = [d_1, d_2, \cdots, d_n]\in\mathbb{C}^{n\times n}$, where
$$ d_l = [1, \omega^{l - 1}, \omega^{2(l - 1)}, \cdots, \omega^{(l - 1)(n - 1)}]^T, l = 1, \cdots, n $$
with $\omega = e^{-2\pi j/n}$ and $j = \sqrt{-1}$.
For simplicity of notation, we consider $a_{n + i} = a_i$.
Then we can verify that
$$ r_id_l = \sum_{k = 1}^{n}{a_{k + i - 1}\omega^{(l - 1)(k - 1)}} = \omega^{-(l - 1)(i - 1)}\sum_{k = 1}^{n}{a_{k}\omega^{(l - 1)(k - 1)}} = (r_1d_l) \cdot d_{li}^* $$
Denote $b_l = r_1d_l$ and we have $Ad_l = b_ld_l^{*}$, where $d_l^{*} = [d_{l1}^{*}, \cdots, d_{ln}^{*}]^{T}$ (not the same as the conjugate transpose $d_l^{H}$).
Similarly, consider $d_l^{*}$ and it is easy to verify that $Ad_l^* = c_ld_l$ where $c_l$ is defined as $c_l = r_1d_l^* = \sum_{k = 1}^{n}{a_{k}\omega^{-(l - 1)(k - 1)}}$.
Now consider $u_l$, $v_l$ such that $u_l^2 = c_l$, $v_l^2 = b_l$ (specified later) and we find
$$ A(u_ld_l + v_ld_l^{*}) = u_lb_ld_l^{*} + v_lc_ld_l = u_lv_l\cdot(u_ld_l + v_ld_l^{*}) $$
Define $\overline{d}_l = u_ld_l + v_ld_l^{*}$ and it is clear that $\overline{d}_l$ is an eigenvector of $A$ and $\lambda_l = u_lv_l$ is the corresponding eigenvalue.
Notice that $d_l$ is one column vector of $D$, and different column vectors of $D$ give eigenvectors of $A$ from the above procedure.
Specially, the all-ones vector $[1, 1, \cdots, 1]^T$ is the first column of $D$, i.e. $d_1$, and $d_1^* = d_1$ so $\overline{d}_1$ is just a multiple of $d_1$, which gives the eigenvector you mentioned in the question.
Now for the second part, let's try prove that all eigenvectors can be obtained from the above procedure.
To begin, it is worth discussion that the definitions of $u_l$ and $v_l$ are actually not unique.
Given that the matrix $A$ is real, then we know that $b_l = c_l^*$.
Without loss of generality, consider $b_l = \rho_l e^{-2j\theta}$ and $c_l = \rho_l e^{2j\theta}$ with $\rho_l\ge 0$, then correspondingly we have $u_l = \pm\sqrt{\rho_l}e^{j\theta}$ and $v_l = \pm\sqrt{\rho_l}e^{-j\theta}$.
To this end, we can see that for every vector $d_l$ we have two possible pairs of eigenvalues and eigenvectors:
$$ \lambda_l^{(1)} = \rho_l, \quad\overline{d}_l^{(1)} = \sqrt{\rho_l}\left(e^{j\theta}d_l + e^{-j\theta}d_l^{*}\right) = 2\sqrt{\rho_l}\cdot\text{Re}(e^{j\theta}d_l) $$
$$ \lambda_l^{(2)} = -\rho_l, \quad\overline{d}_l^{(2)} = \sqrt{\rho_l}\left(e^{j\theta}d_l - e^{-j\theta}d_l^{*}\right) = 2\sqrt{\rho_l}\cdot j\text{Im}(e^{j\theta}d_l) $$
which means there are $2n$ eigenvectors in total.
However, as we soon will see, there are redundancy in these vectors.
First for $d_1$, which is all-ones vector, $b_l = c_l\in\mathbb{R}$, so $\theta = 0$ or $\pi$, $e^{j\theta}d_1\in\mathbb{R}$ and $\overline{d}_1^{(2)} = 0$ which cannot be an eigenvector.
This reasoning is the same for $d_{n/2 + 1}$ (when $n$ is even) which is also a real vector taking values from $\{\pm 1\}$.
For $l\ge 2$, we can verify that $d_l = d_{n + 2 - l}^{*}$, so $\{\overline{d}_l^{(1)}, \overline{d}_l^{(2)}\} = \{\overline{d}_{n+2-l}^{(1)}, \overline{d}_{n+2-l}^{(2)}\}$.
So there are pairs of $d_l$'s that give the same set of two eigenvectors.
On the other hand, if $l\le n/2$,
$$ \rho_l = |b_l| = \left|\sum_{k = 1}^{n}a_k\omega^{(l-1)(k-1)}\right| $$
are different for different $l$'s, so $\{\pm\rho_2, \cdots, \pm\rho_{\lceil n/2\rceil}\}$ are $2(\lceil n/2\rceil - 1)$ distinct eigenvalues of $A$.
Combining the above two paragraphs, we can calculate the number of distinct eigenvalues by: if $n$ is odd, then there are $2(\lceil n/2\rceil - 1) + 1 = n$ distinct eigenvalues; if $n$ is even, then there are $2(n/2 - 1) + 2 = n$ distinct eigenvalues.
So in conclusion, all eigenvalues and eigenvectors can be found from the procedure in the first part, and this completes are proof.
