The (orbital) wave function for $l=1$ doesn't just "have to be" antisymmetric. It demonstrably "is" antisymmetric. The relevant part of this wave function is completely determined, it's a particular function, so we may see whether it's symmetric or antisymmetric and indeed, it's the latter.
Two particles – in this case two pions – orbiting each other are described by a wave function of the relative position
$$\vec r = \vec r_1 - \vec r_2 $$
Writing $\vec r$ in spherical coordinates, a well-defined $l,m$ means that the wave function factorizes to
$$ \psi (\vec r) = \psi_r(r)\cdot Y_{lm}(\theta,\phi)$$
The angular dependence simply has to be given by $Y_{lm}$ because $Y_{lm}$ is, up to the overall normalization, the only wave function with the right angular momentum and its $z$-component being $l,m$.
But $Y_{lm}$ is easily seen to be an odd function of $\hat r$ for odd $l$ and even function of $\hat r$ for even $l$. In particular, $Y_{00}$ is a constant while $Y_{10}$ and $Y_{1,\pm 1}$ are proportional to $z$ and $x\pm iy $ on the unit sphere, respectively.
These functions proportional to $x,y,z$ are clearly odd functions of $\hat r$, so they change the sign under $\vec r\to -\vec r$ which is the sign flip equivalent to $\vec r_1\leftrightarrow \vec r_2$. This odd parity is also called antisymmetry.
In the pion reference frame the two outgoing leptons are very boosted, hence helicity and chirality almost coincide. The angular momentum conservation forces them to have opposite spins, since the pion spin is zero. Therefore, they will have the same helicity, which is highly suppressed in this kinematic regime, because of the vector nature of the QED interactions (see for example Thomson, Modern Particle Physics, chapter 6).
Just my two cents.
Best Answer
As far as I understand, due to conservation of angular momentum, the resulting system of neutral pions would need to have angular momentum 1, therefore, the identical neutral pions would be in an anti-symmetric state, which does not seem possible as they are bosons. Note that a neutral rho meson can decay into two neutral pions and a $\gamma$, although this decay is suppressed.