Here is a simple looking sum which should have an alternate form since it is just a double hypergeometric series with the associated Legendre P function of type 1 $\mathrm P_a^b(z)$
The definitions for the Legendre P function of the first kind uses a regularized Gauss hypergeometric function, Pochhammer symbol, and incomplete beta function:
$$ \begin{split} \mathrm P_v^u(z)=\left(\frac{1+z}{1-z}\right)^\frac u2\, & _2\tilde{ \mathrm F}_1\left(-v,v+1;1-u;\frac{1-z}2\right)\\ &\Downarrow\\ \sum_{n=0}^\infty \mathrm P_{-n}^{-n}(z)=&\sum_{n=0}^\infty \frac{(1+z)^\frac n2\,_2\mathrm F_1\left(1-n,n;n+1;\frac{1-z}2\right)}{(1-z)^\frac n2 n!}\\ =&\sum_{n=0}^\infty \frac{(1+z)^\frac n2}{(1-z)^\frac n2 } \sum_{k=0}^\infty \frac{(1-n)_k(n)_k(1-z)^k}{2^k\Gamma(k+n+1)k!} \\ = &\frac1\pi\sum_{n=0}^\infty \sum_{k=0}^\infty \frac{\sin(\pi n)(k-n)!(1-z)^{k-\frac n2} (1+z)^\frac n2 }{2^k(k+n)k!} \\ =&\sum_{n=0}^\infty \frac{2^n(1-z^2)^\frac n2 \mathrm B_\frac{1-z}2(n,n)}{\Gamma(n)} \end{split}$$
Here is an integral representation:
$$ \begin{split} \mathrm B_z(a,b) = &\int_0^z t^{a-1}(1-t)^{b-1} dt\\ & =\int_{z^{-1}}^\infty t^{-a-b}(t-1)^{b-1} dt\\ &\overset{|z\in\Bbb R|<\infty}{=}\int_0^\infty (t+z^{-1})^{-a-b}(t+z^{-1}-1)^{1-b} dt \\ \implies &\mathrm B_z(a,a) = \int_0^\infty (t+z^{-1})^{-2a}(t+z^{-1}-1)^{1-a} dt \end{split} $$ Therefore we may or may not have: $$ \begin{split} \sum_{n=0}^\infty \mathrm P^{-n}_{-n}(z) &=\sum_{n=0}^\infty \frac{2^n(1-z^2)^\frac n2 \mathrm B_\frac{1-z}2(n,n)}{\Gamma(n)}\\ & = \sum_{n=0}^\infty \frac{2^n(1-z^2)^\frac n2 \int_0^\infty \left(t+ \frac 2{1-z}\right)^{-2n}\left(t+\frac 2{1-z}-1\right)^{1-n} dt }{\Gamma(n)}\\ & =\int_0^\infty \frac{2e^\frac{2\sqrt{1-z^2}}{\left(t+\frac2{1-z}\right)^2\left(t+\frac2{1-z}-1\right)}\sqrt{1-z^2}}{\left(t+\frac2{1-z}\right)^2}dt \end{split} $$ We cannot use an Appell Series since there is a negative coefficient of the sum index in the gamma function argument. Here is an approximated plot:
$\sum\limits_{n=-\infty}^0 \mathrm P_n^n(z)$:
There is a discontinuity at $z=1$. The sum terms themselves are just rational functions of $z$. What is an evaluation of the sum? Please correct me and give me feedback.