Dear Andrew, first, I edited your "anomoly" which should be "anomaly". I couldn't watch it.
Second, all the decays you mention use the weak force and the elementary Feynman diagram is always the same: it's a cubic vertex with one W-boson, one decaying fermion, and one fermionic decay product. So the amplitude is essentially $g_{SU(2)} \bar u_{final} \gamma_\mu u_{initial} \epsilon_W^\mu$.
However, what hugely depends on the mass of the fermions are the kinematic factors - the Lorentz-invariant phase space, if you wish. The "universal" amplitude above has to be integrated over all allowed momenta of the final particles, with the $d^3 p / 2E$ measure. Also, there's $1/2E$ for each initial particle.
The most impressive "failure of dimensional analysis" among these weak decays is the decay rate of the ordinary neutron - its half-life is ten minutes! That's an extremely long time scale, especially if you compare it to the half-life of top quark etc. that you mentioned. Both decays are driven by the same elementary process whose Lorentz-invariant amplitude is essentially identical! The neutron is this stable because it's just slightly heavier than the proton, the main decay product, and the phase space for the allowed electron's and antineutrino's momenta in the final state is just extremely small. (There are probably other similarly long half-lives of unstable nuclei that decay via beta-decay - which are just heavier counterparts of the decaying neutron. The neutron decay is also a simple case of a beta-decay.)
There exists no observed disagreement between any weak decay (of a known particle) and the Standard Model prediction. That's a glimpse of a much more general fact: the Standard Model just universally works. If I am the first one who tells you that it does, it's unfortunate.