Conformal covers of all degrees

Here is a partial answer: If there is such a conformal manifold $M$ of dimension $n\ge 2$, then $M$ admits a flat metric. The reason is that the sequence of conformal covering maps $\phi_k: M\to M$ cannot contain a subsequence converging to a conformal map. Hence, the universal conformal covering $\tilde{M}$ cannot admit a compatible Riemannian metric for which the lifts $\tilde\phi_k: \tilde{M}\to \tilde{M}$ are isometric. Thus, by Ferrand's solution of Lichnerowicz conjecture

Ferrand, Jacqueline, The action of conformal transformations on a Riemannian manifold, Math. Ann. 304, No. 2, 277-291 (1996). ZBL0866.53027.

the manifold $\tilde M$ is either conformal to $S^n$ (which is, of course, impossible) or to $E^n$.

Thus, the problem essentially reduces to the one of flat tori and there should be an algebraic argument proving that $n=1$ in this setting:

Suppose that $\Gamma< Isom(E^n)$ is a discrete cocompact torsion-free subgroup such that the manifold $M=E^n/\Gamma$ admits a covering $\phi: M\to M$ of degree $d$. Then $\phi$ lifts to an affine conformal map $\Phi: E^n\to E^n$. Let $\Lambda< \Gamma$ be the the translation lattice in $\Gamma$. Then $\Phi \Lambda \Phi^{-1}= \Lambda'$ is index $d$ sublattice. In other words, $\Phi$ projects to a degree $d$ conformal self-map $\psi: A\to A$, where $A= E^n/\Lambda$ is a flat torus.

This is the second part of the answer. Suppose $E^n$ is a flat torus admitting a conformal self-map $\varphi_d$ of degree $d$ for every $d=1,2,3,\ldots$. We prove that this is only possible when $n=1$.

Algebraic reformulation: Fix a positive definite symmetric bilinear form $Q$ on $\mathbb{R}^n$, $n\geq 2$. Call an integer $n\times n$ matrix $M$ conformal if $M^t Q M$ is a positive real multiple of $Q$. Degree of such a matrix is $\det M$. We prove that it is not possible to have a conformal matrix of degree $d$ for each $d=1,2,3,\ldots$. Suppose the contrary, i.e. there is such an integer matrix $M_d$ for every $d$.

First, by taking determinants for every $d$ we find the coefficient of proportionality $$ M_d^t Q M_d= d^{2/n} Q. $$ Normalize $Q$ so that $Q_{11}=1$. For any vectors $u,v$ denote $(u,v)=u^t Q v$. Let $v_d$ be the first column of $M_d$. Then we have $$ (v_d, v_d) = d^{2/n}\qquad (d=1,2,3,\ldots) $$ We claim that this is impossible. Consider the case $n=2$ first. Note that no two among $v_1, v_2, v_3$ can be collinear. Hence $v_3=\alpha v_1 + \beta v_2$ for some $\alpha, \beta\in\mathbb{Q}$. This allows to compute all the entries of $Q$ out of $\alpha, \beta$ and deduce that they are rational. So we have $a,b,c\in\mathbb{Q}$ so that the equation $a x^2 + b xy + c y^2=d$ has solutions in integers for every $d$, but $b^2-4a c<0$. This is impossible: by Chebotarev density theorem one can choose a prime $p$ such that $p$ doesn't divide the numerators and the denominators of $a,b,c$ and the equation $a x^2 + b x + c=0$ has no roots mod $p$. Setting $d=p$ leads to a contradiction.

Now consider the case $n\geq 3$. Consider the numbers of the form $p^{2/n}$ for prime numbers $p>n$. They are linearly independent over $\mathbb{Q}$ because the field extension generated by $p^{2/n}$ is ramified at $p$, and can only be further ramified at the divisors of $n$. On the other hand, consider the sequence of integer $n\times n$ matrices $v_p v_p^t$. We have an infinite sequence of elements of a finite dimensional vector space, so there must be a linear relation $$ \sum_{i=1}^N c_i v_{p_i} v_{p_i}^t = 0 \qquad ((c_1,c_2,\ldots,c_N)\in\mathbb{Q}^N\setminus \{0\}) $$ for prime numbers $p_1,p_2,\ldots,p_N>n$. This implies $$ \sum_{i=1}^N c_i p_i^{2/n} = \sum_{i=1}^N c_i (v_{p_i}, v_{p_i}) = \sum_{i=1}^N c_i \operatorname{trace}(Q v_{p_i} v_{p_i}^t)=0, $$ a contradiction.