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 $M_{red} = H^{-1}(E)/{\lbrace\textrm{Hamiltonain flow}\rbrace}$. Since the unreduced action acts transitively on $H^{-1}(E)$, the reduced action acts transitively on $M_{red}$. The $\operatorname{SO}(n)$-equivariance of the corresponding reduced momentum map $J_{red}:M_{red}\rightarrow \mathfrak{so}(n)^*$ implies that $J_{red}$ maps $M_{red}$ to a coadjoint orbit $\mathcal{O}\subset\mathfrak{so}(n)^*$. Kostant's coadjoint orbit covering theorem then implies that $J_{red}$ is a symplectic covering map of $M_{red}$ 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 $J_{red}$ 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 $\pi:H^{-1}(E)\rightarrow M_{red}$ and the momentum map $j:H^{-1}(E)\rightarrow \mathcal{O}$ agree, which implies that the covering $j_{red}:M_{red}\rightarrow \mathcal{O}$ 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 self-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 $\ker(\operatorname{ad}_\xi) = \mathfrak{so}(n)_\xi$, and we can choose a Cartan subalgebra $\mathfrak{h}\subset\mathfrak{so}(n)_\xi$, from which you can define 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 :)).