Another way to view this: think of $T^*S^{n-1}$ as $\lbrace (q,p)\in\mathbb{R}^{2n}\,|\,\Vert q\Vert = 1, q.p=0\rbrace$, with symplectic structure $\omega((v_q,v_p),(w_q,w_p)) := v_q\cdot w_p - v_p\cdot w_q$. The  $\operatorname{SO}(n)$-action is Hamiltonian, with $\operatorname{SO}(n)$-equivariant momentum map $J:T^*S^{n-1}\rightarrow \mathfrak{so}(n)^*$ given by
$$
\langle J(q,p),\xi\rangle = p(\xi\cdot q)\qquad\textrm{for}\qquad\xi\in\mathfrak{so}(n).
$$

Since the $\operatorname{SO}(n)$-action preserves the Hamiltonian, it drops to the reduced space $M_{\textrm{red}} = H^{-1}(E)/{\lbrace\textrm{Hamiltonian flow}\rbrace}$. Since the unreduced action acts transitively on $H^{-1}(E)$, the reduced action acts transitively on $M_{\textrm{red}}$. The $\operatorname{SO}(n)$-equivariance of the corresponding reduced momentum map $J_{\textrm{red}}:M_{\textrm{red}}\rightarrow \mathfrak{so}(n)^*$ 
implies that $J_{\textrm{red}}$ maps $M_{\textrm{red}}$ to a coadjoint orbit $\mathcal{O}\subset\mathfrak{so}(n)^*$. [Kostant's coadjoint orbit covering theorem][1] then implies that $J_{\textrm{red}}$ is a symplectic covering map of $M_{\textrm{red}}$ onto $\mathcal{O}$
, equipped with the (positive) Kostant-Kirillov-Souriau form
$$
\omega^+_\mu(-\operatorname{ad}_\zeta^*\mu,-\operatorname{ad}_\chi^*\mu) := \langle\mu,[\zeta,\chi]\rangle.
$$
In fact, in this case $J_{\textrm{red}}$ is a symplectomorphism.

To see things more concretely: identifying $\mathfrak{so}(n)$ with its dual via the trace form, we get the (still equivariant) Lie-algebra-valued momentum map $j:T^*S^{n-1}\rightarrow \mathfrak{so}(n)$ given by
$$
j(q,p) = \frac{1}{2}(q p^\top - p q^\top)\in\mathfrak{so}(n).
$$
Then for example taking the generic point $(q,p) = (e_1,\sqrt{2E} e_2)\in H^{-1}(E)$, this gives
$$
j(q,p) = \begin{bmatrix} 0 & \sqrt{\frac{E}{2}} & \ldots & 0\\
-\sqrt{\frac{E}{2}} & 0 &\ldots & 0 \\
\vdots \\
0 & 0 &\ldots & 0\end{bmatrix}
$$
Clearly the stabiliser of this element under the adjoint action of $\operatorname{SO}(n)$ is $\operatorname{SO}(2)\times\operatorname{SO}(n-2)$. Viewed in $T^*S^{n-1}$, the second factor acts trivially on $(q,p)$, while the first factor produces the great circles corresponding to the Hamiltonian flow. Hence the fibres of the quotient map $\pi:H^{-1}(E)\rightarrow M_{\textrm{red}}$ and the momentum map $j:H^{-1}(E)\rightarrow \mathcal{O}$ agree, which implies that the covering $j_{\textrm{red}}:M_{\textrm{red}}\rightarrow \mathcal{O}$ is actually a diffeomorphism.

It's straightforward now to put a complex structure on the coadjoint/adjoint orbit (the same construction works for the orbits of any compact Lie group). I will use adjoint orbits here. At any point $\xi\in \mathcal{O}\subset \mathfrak{so}(n)$, the operator $\operatorname{ad}_\xi:\operatorname{so}(n)\rightarrow\operatorname{so}(n)$ is skew-adjoint with respect to the trace form, and so has pure imaginary eigenvalues. Let 
$$
\mathfrak{n}_\xi^+ := \lbrace \zeta\in\mathfrak{so}(n)\,|\, \operatorname{ad}_\xi\zeta = i\lambda\, \zeta \quad\textrm{for some }\lambda>0\rbrace,
$$
and define the polarization $F$ on $\mathcal{O}$ by
$$
F_\xi := \lbrace \operatorname{ad}_\zeta \xi\,|\,\zeta\in\mathfrak{n}_\xi^+\rbrace.
$$
At any point $\xi\in\mathcal{O}$, we have $\ker(\operatorname{ad}_\xi) = \mathfrak{so}(n)_\xi$ (the adjoint stabiliser algebra), and we can choose a Cartan subalgebra $\mathfrak{h}\subset\mathfrak{so}(n)_\xi$, from which we can introduce an ordering on weights. As Ben McKay mentioned, the usual Borel-Weil construction implies that the holomorphic sections with respect to the polarization $F$ give an irreducible representation with highest weight related to $\xi$. I would have to check my signs and conventions carefully, but I think in this case you get the dual to the irrep with highest weight $-\frac{i}{\hbar}\xi^\flat$, where $\cdot^\flat:\mathfrak{so}(n)\rightarrow \mathfrak{so}(n)^*$ is the duality with respect to the trace form (don't quote me on that, though :)).


  [1]: https://books.google.co.uk/books?id=k-7kBwAAQBAJ&pg=PA463&lpg=PA463&dq=kostant%27s+coadjoint+covering+theorem&source=bl&ots=yVUr6cUfpP&sig=nc8UJzA_JhcIcoaFL67LdwbY1o8&hl=en&sa=X&ved=0ahUKEwj0poDovNjYAhXEzqQKHdkABaAQ6AEIMzAC#v=onepage&q=kostant's%20coadjoint%20covering%20theorem&f=false