The harmonic trap gives an energy dependent effective volume. Think in classical terms: at energy $E$, the accessible region for the particle is a ball of radius $r$ and volume $V$ with:
\begin{align}
r &= \sqrt{\frac{2E}{m\omega^2}} \\
V &= \frac{4\pi}{3}\left(\frac{2E}{m\omega^2}\right)^{3/2}
\end{align}
Adding thermal fluctuations, in the canonical ensemble, the spacial distribution is Gaussian with root mean square distance $r_{rms}$ (given by the equipartition theorem):
$$
r_{rms} = \sqrt{\frac{3k_BT}{m\omega^2}}
$$
which gives an effective accessible ball of volume:
$$
V = \frac{4\pi}{3}\left(\frac{3k_BT}{m\omega^2}\right)^{3/2}
$$
Up to a numerical factor, this is the effective volume you find in the numerator of your expression. This is because $r_{rms}$ is the only length scale of your problem, so by dimensional analysis a volume will necessarily proportional to it's cube.
In short, you should think of $1/\omega$ as the effective length of your box. In particular, just as for the gas in a box, you have the same extensive scaling. Since $V\sim \omega^{-D}\sim N$, you need $\omega\sim N^{-1/D}$. The temperature dependence is there in some sense to match the dimensions.
For your second question, the temperature dependence is purely classical. As temperature increase, thermal fluctuations increase, the "available" energy increases which increases the effective volume by the first equations.
Note that the purely classical interpretation works above the critical temperature. Below the critical temperature, you'll have a macroscopic number of bosons in the ground state. Their effective occupied volume will be given by the ground state of the harmonic oscillator. It is therefore temperature independent:
$$
r_{rms} = \sqrt{\frac{\hbar}{2m\omega}} \\
V = \frac{4\pi}{3}\left(\frac{\hbar}{2m\omega}\right)^{3/2}
$$
Thus, in the BEC phase, the classical fraction will still have a temperature decreasing effective volume, but the ground state fraction will have a fixed volume just like for a gas in a box.
Btw, if you wanted a full quantum treatment, this would modify the total particle number to:
$$
N = \sum_{n\in\mathbb N^3} \frac{ze^{\beta \hbar \omega(n_x+n_y+n_z)}}{1-ze^{\beta \hbar \omega(n_x+n_y+n_z)}}
$$
It is consistent with the classical treatment for all temperature because in the thermodynamic limit, if the temperature is intensive, $\beta\omega\hbar \gg 1$. You just need to treat apart from the ground state. The classical theory applies in particular the equipartition theorem as well.
Hope this helps.
