Consider the differential equation for the associated Legendre polynomials, $(1-z^2)w''(z) - 2z w(z) + \left(\nu(\nu+1) - \frac{\mu^2}{1-z^2}\right)w(z) = 0.$ Change variables. Let $z = \cos \frac{x}{\nu}$. (Notice, for example, that $\frac{d}{dz} = -\frac{\nu}{\sin\frac{x}{\nu}} \frac{d}{dx}$.) In the limit $\nu\to\infty$ the DE takes the form $x^2 w''(x) + x w'(x) + (x^2-\mu^2)w(x) = 0$ which is, of course, Bessel's equation. Therefore,
$\lim_{\nu\to\infty} P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right) \propto J_\mu(x).$ Since it is getting late, I leave it as an exercise to find the constant.
Addendum: The argument above tells us that in the limit $P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right)$ is some combination of solutions to Bessel's equation. The singular solution $Y_\mu(x)$ is ruled out since $P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right)$ is not singular at $x=0$.
Using the integral representation for $-1 and $\mathrm{Re}\,\mu > 0$, $P_\nu^{-\mu}(z) = \frac{(1-z^2)^{-\mu/2}}{\Gamma(\mu)} \int_z^1 d t\, P_\nu(t)(t-z)^{\mu-1},$ we find for $x\ll 1 \ll \nu$ that $P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right) \sim \frac{1}{\Gamma(\mu+1)} \left(\frac{x}{2\nu}\right)^\mu.$ (Here we exploit the fact that for $x\ll 1$, $P_\nu\left(\cos \frac{x}{\nu}\right) = 1+O(x^2)$.) But for small $x$ we have $J_\mu(x) \sim \frac{1}{\Gamma(\mu+1)} \left(\frac{x}{2}\right)^\mu$ and so $\lim_{\nu\to\infty} \left[\nu^\mu P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right)\right] = J_\mu(x).$