A nice overview of the problem is given in arXiv:1205.3740. I'll summarise the most important points here.
Let $d$ be the number of space dimensions. Then the Laplace operator is given by
$$
\Delta=\frac{\partial^2}{\partial r^2}+\frac{d-1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\Delta_S\tag{1}
$$
where $\Delta_S$ is the Laplace operator on the $d-1$ sphere.
The Coulomb potential is given by the solution of
$$
-\Delta V=e\delta(\boldsymbol r)\tag{2}
$$
which is solved by
$$
V(r)=\frac{2(d/2)!}{(d-2)\pi^{(d-2)/2}}\frac{e}{r^{d-2}}\tag{3}
$$
One way to show the expression above is to consider the Gauss Law in $d$ dimensions, that is, $E(r)=e/\int_r\mathrm ds$ where the area of the $d-1$ sphere can be found here.
With this, the Schrödinger equation reads
$$
\left[\frac{1}{2m}(-\Delta)+V(r)-E\right]\psi(\boldsymbol r)=0\tag{4}
$$
The problem has spherical symmetry so that, as usual, we may write
$$
\psi(r,\boldsymbol \theta)=\frac{1}{r^{(d-1)/2}}u(r)Y_\ell(\boldsymbol\theta)\tag{5}
$$
where $\boldsymbol \theta=(\theta,\phi_1,\phi_2,\cdots,\phi_{d-2})$ and the power $(d-1)/2$ of $r$ is chosen to remove the linear term in $(1)$. The superspherical harmonics ($\sim$ Gegenbauer polynomials) are the generalisation of the usual spherical harmonics to $d$ dimensions:
$$
\Delta_S Y_\ell+\ell(\ell+d-2)Y_\ell=0\tag{6}
$$
Using this form for $\psi$, the Schrödinger equation becomes
$$
u''(r)+2m[E-V_\ell(r)]u(r)=0\tag{7}
$$
where the effective potential is
$$
V_\ell=V(r)+\frac{1}{2m}\frac{\ell_d(\ell_d+1)}{r^2}\tag{8}
$$
with $\ell_d=\ell+(d-3)/2$.
Now, this eigenvalue problem has no known analytical solution, so that we must resort to numerical methods. You can find the numerical values of the energies inon the arxiv article. An important point that is addressed in that article is that there are no negative eigenvalues for $d\ge 4$, that is, there are no bound states in more than three dimensions. But for $d\ge 5$ there are stable orbits, with positive energy, with well behaved wave functions.
To answer to some of your questions:
In general you need $d$ quantum numbers for $d$ dimensions, modulo degeneracies. In the case of the hydrogen atom in 3D, there are the spherical symmetry and the accidental symmetry, so that you only need one quantum number. In $d$ dimensions the spherical symmetry remains, but I think that the accidental does not, which would mean that you need $d-1$ quantum numbers for $d\neq 3$. One would need to check whether the Runge-Lenz vector is conserved in $d$ dimensions or not (left as an exercise). If this vector is actually conserved, then the energies would depend on $d-2$ quantum numbers.
As there is no analytical solution to the radial Schrödinger equation, we don't know. In the case of $d=3$ dimensions, the Bohr-Sommerfeld quantisation rule turns out to be exact. We could check what this scheme predicts for $E_n$ (though we couldn't know whether it would be exact or not: we must compare to the numerical results).
These are well-known for mathematicians. You can read about them on the wikipedia article.
In closed form, no. I don't know of an asymptotic formula, but it should be rather easy to derive from the radial Schrödinger equation, where the centrifugal term $r^{-2}$ dominates for $r\to\infty$, so that we can neglect the Coulomb term.
