It is perfectly alright to have (associated) Legendre functions of half-integer order (or in fact, any arbitrary complex order). The key is to define them in terms of Gaussian hypergeometric functions, e.g.
$P_\ell^m(z)=\frac{(1+z)^{m/2}}{(1-z)^{m/2}}\frac1{\Gamma(1-m)}{}_2 F_1\left({{-\ell\quad\ell+1}\atop{1-m}} \mid \frac{1-z}{2}\right)$
to use one particular normalization.
For $\ell$ a half integer, one could use the formulae
$\begin{align*} P_{-\frac12}^m(z)&=\frac2{\pi}K\left(\frac{1-z}{2}\right)\\ P_\frac12^m(z)&=\frac2{\pi}\left(2E\left(\frac{1-z}{2}\right)-K\left(\frac{1-z}{2}\right)\right) \end{align*}$
where $K(m)$ and $E(m)$ are complete elliptic integrals with parameter $m$, along with the usual recursion relations over $\ell$ and $m$.
For half-integer $m$ we can start the recursion relations with
$\begin{align*} P_0^{-\frac12}(z)&=\frac2{\sqrt\pi}\sqrt[4]{\frac{1-z}{1+z}}\\ P_0^\frac12(z)&=\frac1{\sqrt\pi}\sqrt[4]{\frac{1+z}{1-z}} \end{align*}$
and for both $\ell$ and $m$ half-integer, the recursion relations take the initial values
$\begin{align*} P_{-\frac12}^{-\frac12}(z)&=2\sqrt{\frac2{\pi}}\frac1{\sqrt[4]{1-z^2}}\arcsin\left(\sqrt{\frac{1-z}{2}}\right)\\ P_{-\frac12}^\frac12(z)&=\sqrt{\frac2{\pi}}\frac1{\sqrt[4]{1-z^2}}\\ P_\frac12^{-\frac12}(z)&=\sqrt{\frac2{\pi}}\sqrt[4]{1-z^2}\\ P_\frac12^\frac12(z)&=\sqrt{\frac2{\pi}}\frac{z}{\sqrt[4]{1-z^2}} \end{align*}$
Note that for arbitrary order, the Rodrigues formula can be suitably interpreted as a Riemann-Liouville differintegral (i.e., differentiation/integration to arbitrary complex order). I might edit this answer for a demonstration later.