I find it useful to represent the Legendre function in terms of a hypergeometric function, using a formula from Wikipedia,

$$f(\theta)=\frac{ (1+\cos \theta)^{\mu/2} \, _2F_1\left(-\nu,\nu+1;1-\mu;\frac{1}{2} (1-\cos \theta)\right)}{\Gamma (1-\mu)\sin^\mu\theta(1-\cos \theta)^{\mu/2}}.$$

Then I find the small-$\theta$ expansion $$f(\theta)=\frac{ 4 (\mu-1)+\tfrac{1}{3}\theta^2 \bigl((1-\mu) \mu-3 \nu (\nu+1)\bigr)+{\cal O}(\theta^4)}{2^{2-\mu}\theta^{2\mu} \Gamma (2-\mu)}.$$ The leading order term is $\nu$-independent, as noted in the OP, the next order term does depend on $\nu$.

I find it useful to represent the Legendre function in terms of a hypergeometric function, using a formula from Wikipedia,

$$f(\theta)=\frac{ (1+\cos \theta)^{\mu/2} \, _2F_1\left(-\nu,\nu+1;1-\mu;\frac{1}{2} (1-\cos \theta)\right)}{\Gamma (1-\mu)\sin^\mu\theta(1-\cos \theta)^{\mu/2}}.$$

Then Mathematica gives me the small-$\theta$ expansion $$f(\theta)=\frac{ 4 (1-\mu)+\theta^2 \bigl(\tfrac{1}{3}(1-\mu) \mu- \nu (\nu+1)\bigr)+{\cal O}(\theta^4)}{2^{2-\mu}\theta^{2\mu} \Gamma (2-\mu)}.$$ The leading order term is $\nu$-independent, as noted in the OP, the next order term does depend on $\nu$.