Clausen’s identity for Legendre polynomials has the form (see, for example, A generating function of the squares of Legendre polynomials, by Wadim Zudilin: https://arxiv.org/abs/1210.2493) $$P_n(\cos{\theta})^2=\sum_{k=0}^n\frac{(-1)^k}{2^{2k}}\binom{n}{k}\binom{n+k}{n}\binom{2k}{k}\sin^{2k}{\theta}.$$ Do the analogous identity exist for the square of associated Legendre polynomials $[P_n^l(\cos{\theta})]^2$?

There is a similar formula
$$
\small{\left(P_n^m(\cos\theta)\right)^2=(\sin{\theta})^{2m}\frac{(m+n)!}{(n-m)!}\sum_{k=0}^{n-m}\frac{(-1)^k}{4^{k+m}}\binom{n+m}{k+2m}\binom{n+k+m}{n+m}\binom{2k+2m}{k+m}\sin^{2k}{\theta}}
$$
obtained from the following representation for associated Legendre polynomials
$$
P_n^m(z)=(-1)^m \left(\frac{1-z}{1+z}\right)^{\frac{m}{2}} \frac{(m+n)!}{m! (n-m)!} \, _2F_1\left(-n,n+1;m+1;\frac{1-z}{2}\right)
$$
and the formula (6.1) from the paper W.N. Bailey, *Some Theorems Concerning Products of Hypergeometric Series*, Proc. London Math. Soc. **38**, 377–384 (1935):
$$
{}_2F_1(a,b;c;x){}_2F_1(a,b;a+b-c+1;x)={}_4F_3\left({a,b,\tfrac{a+b}2,\tfrac{1+a+b}2\atop a+b,c,a+b-c+1};4x(1-x)\right).
$$