Why doesn't aliasing of broad band noise 'pile up' in the sample band?
You are correct: after sampling, the aliased noise components do pile up in the frequency band below the Nyquist frequency. The question is just what exactly it is that piles up, and what is its consequence.

In the following I assume that we deal with random noise modeled as a wide-sense stationary (WSS) random process, i.e. a random process for which we can define a power spectrum. If \$N(t)\$ is the noise process and \$R_k= N(kT)\$ is the sampled noise process (with sample period \$T\$), then the power spectrum of \$R_k\$ is an aliased version of the power spectrum of \$N(t)\$:

$$S_R(f)=f_s\sum_{k=-\infty}^{\infty}S_N(f-kf_s)\tag{1}$$

where \$f_s=1/T\$ is the sampling frequency. Of course, if \$N(t)\$ is band-limited (which is always the case) then only a finite number of shifted power spectra of \$N(t)\$ add up in the band of interest \$[0,f_s/2]\$.

The noise power is given by the integral of the respective power spectrum. In the case of \$N(t)\$ we have to integrate over the whole bandwidth of \$N(t)\$, whereas in the case of the sampled noise \$R_k\$ we have to integrate in the band \$[0,f_s/2]\$. From (1) it becomes clear that in both cases we obtain the same power because either we integrate the original power spectrum \$S_N(f)\$, or we integrate an aliased (i.e., piled up) version in the band \$[0,f_s/2]\$.


  Consequently, the noise power does not change after sampling, regardless of the sampling frequency. The sampled noise has the same power as the original continuous-time noise.


So the power of the sampled noise only changes if you change the power of the continuous-time noise, and this can be done by the anti-aliasing filter, because the filter reduces noise band-width and, consequently, noise power. Note that only looking at the peak-to-peak value doesn't say much, because you need to consider the power.


Reference:

E.A. Lee, D.G. Messerschmitt: Digital Communication, 2nd ed., section 3.2.5 (pp. 64)