Summary: this is a topic of active research [1,2,3,4]; the answer to the general case is not known; answers exist for $a\in\mathbb{C}$, $b,c,\in\mathbb{N}$, and for $a,b,c\in\mathbb{C}$ with $b=c$.
Case 1: Consider first the case that one of the three parameters $a,b,c$ is complex, while the other two are still integers. For the Legendre function I use the representation
$$P_\mu(z)=\, _2F_1\left(-\mu,\mu+1;1;\frac{1-z}{2}\right),\;\;|1-z|<2,$$
and for the 3j symbol squared
$$\left(\begin{array}{ccc}
a & b & c\\
0 & 0 & 0
\end{array}\right)^{2}=\frac{\binom{a+b-c}{\frac{1}{2} (a+b-c)} \binom{a-b+c}{\frac{1}{2} (a-b+c)} \binom{-a+b+c}{\frac{1}{2} (-a+b+c)}}{(a+b+c+1) \binom{a+b+c}{\frac{1}{2} (a+b+c)}}.$$
Then
$$\int_{-1}^1 P_{a}(x)P_{b}(x)P_c(x)\,dx=2\cos ^2\left(\tfrac{1}{2} \pi (a+b+c)\right)\left(\begin{array}{ccc}
a & b & c\\
0 & 0 & 0
\end{array}\right)^{2},$$
$$\text{for}\;\;a\in\mathbb{C},b\in\mathbb{N},c\in\mathbb{N}.$$
(I checked this with Mathematica, which can evaluate the integral in closed form.)
So in this case of one complex variable the function $f(a,b,c)$ in the OP equals $2\cos ^2\left(\tfrac{1}{2} \pi (a+b+c)\right)$. Note that if all three variables are integers, the $\cos^2$ factor can be set to unity, since the 3j symbol equals zero for $a+b+c$ odd.
Case 2: If all three parameters are complex, a closed form expression in terms of generalized 3j symbols (or equivalently, generalized Clebsch-Gordan coefficients) exists for the integral
$$T_{\mu,\nu}=\int_{-1}^1 P_{\mu}(x)P_{\nu}(x)P_\nu(-x)\,dx,\;\;\mu,\nu\in\mathbb{C}.$$
Note the sign change of one argument of the Legendre function. For integer $\nu=n$ one would simply have $P_n(-x)=(-1)^nP_n(x)$, but for noninteger $\nu$ the sign change is essential.
The result, derived by Yajun Zhou [1] (see also [2,3,4]) is given below, it is lengthy. A simple case is $\mu=\nu$, when we have
$$\int_{-1}^1 P_{\nu}(x)P_{\nu}(x)P_\nu(-x)\,dx=\frac{1}{3} \bigl(2+3 \cos \pi \nu+\cos 2 \pi \nu\bigr)\left(\begin{array}{ccc}
\nu & \nu & \nu\\
0 & 0 & 0
\end{array}\right)^{2},\;\;\nu\in\mathbb{C}.$$
For integer $\nu$ the integral vanishes for $\nu$ odd, and for $\nu$ even the usual formula is recovered.
Here is the general result:
\begin{align}
T_{\mu,\nu}={}&\frac{2}{\pi^2}\frac{\sin(\mu\pi)\sin(\nu\pi)}{\mu(\mu+1)}\left[ \, \frac{1}{\nu}\,{_4F_3}\left(\left.\begin{array}{c}
1,\frac{1-\mu}{2},\frac{\mu+2 }{2},-\nu \\[4pt]
\frac{2-\mu }{2},\frac{\mu+3 }{2},1-\nu \\
\end{array}\right| 1\right) -\frac{1}{\nu+1} \,{ _4F_3}\left(\left.\begin{array}{c}
1,\frac{1-\mu}{2},\frac{\mu+2 }{2},\nu+1\ \\[4pt]
\frac{2-\mu }{2},\frac{\mu+3 }{2},\nu+2 \\
\end{array}\right| 1\right) \right].
\end{align}
The combination of two hypergeometric functions ensures the symmetry $T_{\mu,\nu}=T_{-\mu-1,\nu}=T_{\mu,-\nu-1}$, required by $P_{\mu}(z)=P_{-\mu-1}(z)$.
- Yajun Zhou, Legendre
Functions, Spherical Rotations, and Multiple Elliptic Integrals
- Marco Cantarini, A
note on Clebsch-Gordan integral, Fourier-Legendre expansions and
closed form for hypergeometric series
- John Campbell, New Clebsch-Gordan-type integrals involving threefold products of complete elliptic integrals
- Yajun Zhou, On Some Integrals Over the Product of Three Legendre Functions
