for large $n$, any finite set $\{\nu_1,\nu_2,\ldots \nu_k\}$ becomes a set of i.i.d. Gaussians with zero mean and variance $1/n$, so
$$E\left[e^{\alpha\nu_1}\right]=\sqrt{\frac{n}{2\pi}}\int_{-\infty}^\infty dx\, e^{\alpha x}e^{-nx^2/2}=e^{\alpha^2/2n}.$$
(there is a factor of two difference in the exponent with respect to the OP)

to see the Gaussian limit, start from
$$
P(\nu_1,\nu_2,\ldots \nu_n)\propto\delta\left(1-\sum_{i=1}^{n}\nu_i^{2}\right).
$$
Integrate
out $n-k$ elements, to arrive at the distribution
$$P(\nu_1,\nu_2,\ldots \nu_k)\propto\left(1-\sum_{i=1}^{k}\nu_i^{2}\right)^{(n-k-2)/2} \theta\left(1-\sum_{i=1}^{k}\nu_i^{2}\right),$$
with $\theta(x)$ the unit step function. In the large-$n$ limit at fixed $k$ this tends to
$$P(\nu_1,\nu_2,\ldots \nu_k)\propto\exp\left(-\frac{n}{2}\sum_{i=1}^k \nu_i^2\right)$$

The new version of the question now asks for an $n$-dependent $\alpha$; then one will just have to work with the exact expression,
$$P(\nu_1)=C_n\left(1-\nu_1^{2}\right)^{(n-3)/2} \theta\left(1-\nu_1^{2}\right),\;\;C_n=\frac{2 \Gamma \left(\frac{n}{2}\right)}{\sqrt{\pi } \,\Gamma \left(\frac{n-1}{2}\right)}$$
and hence
$$E\left[e^{\alpha_n\nu_1}\right]=2^{\frac{n}{2}-1} \alpha_n^{1-\frac{n}{2}} \Gamma \left(\frac{n}{2}\right) \left(L_{\frac{n}{2}-1}(\alpha_n)+I_{\frac{n}{2}-1}(\alpha_n)\right)$$
with $I$ a Bessel function and $L$ a Struve function.