This is because the chiral anomaly in the gauge theory with fermions can appear only when there are:
The diagrams diverging worse than logarithmically (i.e., linearly, quadratically and so on);
The diagrams containing an odd number of $\gamma_{5}$ matrices (in another case it is always possible to find the regularization conserving the symmetry).
The diverging diagram containing two currents doesn't contain an odd number of $\gamma_{5}$ matrices. This can be heuristically understood since in the lowest order of the perturbation theory it contains the trace
$$
\tag 1T_{\mu\nu}^{ab}(k) \sim \int d^{4}p\text{tr}\big[D(p)\gamma_{\mu}t_{a}\gamma_{5}D(p -k)\gamma_{\nu}t_{b}\gamma_{5}\big],
$$
where $t_{a}$ is the symmetry generator with the color index $a$, $D(p)$ is the fermion propagator, and $k$ is external momentum. The fact that $\gamma_{5}$ matrix can be eliminated from $(1)$ follows from anti-commutation law $\{\gamma_{\mu},\gamma_{5}\}_{+} = 0$ (note that You need to be careful with this relation when using some regularizations).
In its turn, the diverging diagram containing three chiral currents contains unremovable $\gamma_{5}$ matrix, since the corresponding vertex has the form
$$
T_{\mu\nu \lambda}^{abc}(k_{1},k_{2}) \sim \int d^{4}p\text{tr}[D(p+k_{1})\gamma_{\mu}t_{a}\gamma_{5}D(p +k_{1}-k_{2})\gamma_{\nu}t_{b}\gamma_{5}D(p)\gamma_{\lambda}t_{c}\gamma_{5}],
$$
and the $\gamma_{5}$ remains.