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
$$
V(r)=\frac{2(d/2-1)!}{(d-2)\pi^{(d-2)/2}}\frac{e}{r^{d-2}}\tag{3}
$$
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 on 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$^1$, 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.
$^1$ In spherical symmetric systems, the potential depends on neither $\theta$ nor $\phi$. In these systems energies don't depend on $m$, the azimuthal quantum number. Therefore, in general for radial potentials the energies depend on two quantum numbers, $n,\ell$. In the specific case of $V=1/r$, there is another symmetry that is kind of unexpected (or at least it is not very intuitive geometrically). When $V=1/r$ the rotational symmetry $SO(3)$ is enlarged into a $SO(4)$ symmetry, and this new symmetry is known as accidental symmetry. This symmetry makes the energy levels independent of $\ell$, that is, $E=E_n$. Note that this symmetry is not present in the rest of the atomic table, that is, multielectronic atoms, which makes the energy levels depend on the angular momentum (and thus, Hund's rules).
One may illustrate the above as
\begin{aligned}
\text{If $V=V(r,\theta,\phi)$} &\longrightarrow E=E_{n\ell m}\\
\text{If $V=V(r)$}&\longrightarrow E=E_{n\ell}\\
\text{If $V=1/r$}&\longrightarrow E=E_{n}
\end{aligned}
The first line is the general result for 3D systems; the second line is the result of spherical symmetry; and the third line is the result of the accidental symmetry. If you want to read more about this symmetry, you'll find some nice references in Why are hydrogen energy levels degenerate in $\ell$ and $m$? and/or here.
Best Answer
A "Bohr orbit" is related to a classical orbit via the correspondence principle, but not all sets of quantum numbers for hydrogen-like atoms correspond to Bohr orbits. Quantum mechanics is richer than Bohr initially imagined.
Let's consider a hydrogen-like atom with quantum numbers $(n,\ell,m)$. The radial part of the wavefunction is
$$ R_{n\ell}(r) = \sqrt{\text{stuff}}\times e^{-u/2} u^\ell L_{n-\ell-1}^{2\ell+1}(u) \quad\quad \text{where } u = \frac{2Zr}{n a} $$
where $r$ is the radial coordinate, $Z$ the nuclear charge, and $a$ the Bohr radius. The $L_\alpha^\beta$ are the associated Laguerre polynomials, which are polynomials of order $\alpha$. So for an orbital with maximal angular momentum $l=n-1$, the Laguerre stuff is just a constant, and the radial wavefunction $R\sim e^{-r} r^\ell$ has a zero at the origin (for nonzero $\ell$) and a single maximum at some finite $r$. For large $n,\ell$ this single peak is narrow and it makes sense to think of the electron as being "radially localized," like a particle in a circular orbit.
Similarly, the spherical harmonics are given by an associated Legendre polynomial in the variable $\cos\theta$, multiplied by an azimuthal phase $e^{im\phi}$. The extremal Legendre polynomials all have the form
\begin{align} P_\ell^\ell (x) &= (\text{constant}) (1-x^2)^{\ell/2} \\ P_\ell^\ell (\cos\theta) &= (\text{constant}) \sin^{\ell}\theta \end{align}
So for a hydrogen electron where the angular momentum projection is maximized, $|m|=\ell$, the angular probability distribution is strongly peaked at the equator; that peak gets narrower for larger $\ell$. The complex phase increases linearly as you go around in $\phi$.
Taken together, an electron in a hydrogen-like orbital with maximum angular momentum and maximum angular momentum projection onto the $z$-axis, with quantum numbers $(n,\ell,m) = (\ell+1,\ell,\pm\ell)$, has its probability concentrated in a narrow ring around the equator of the coordinate system, with the phase changing around the ring. If you include the time-dependent part of the wavefunction, multiplying by $e^{-i\omega t}$, you have a phase change with time which you can use to find a "probability current." This is the Schrodinger version of a Bohr orbit.
If $\ell$ is large enough that the Bohr orbit is a good description, then the atom can in fact radiate —— not continuously, but by emitting photons and going to orbits with smaller $n$, and eventually to the ground state.
Beware of visualizations like the orbitron, linked in your post. For the case of the $p$-orbitals ($\ell=1$), chemists like to refer to the spatially-oriented $p_x$ and $p_y$, which are rotated versions of $p_z$. However, $p_z$ has definite $m=0$; the $m=\pm1$ states are linear combinations of $p_x \pm i p_y$. For the higher $\ell$, there are more choices to make. I think the chemists' approach is to combine wavefunctions with equal-magnitude $m$, so that the combined wavefunctions are real.