I want to calculate the integral $$\int_{\mathbb{R}^N \times \mathbb{R}^N} \chi_{[0,E]}\left(\sum_{i=1}^N \frac{p_i^2}{2m} + \frac{m \omega^2 q_i^2}{2} \right) \,dp\, dq.$$
Now I should explain what I mean by this. So $p = (p_1,\ldots,p_N)$ and $q = (q_1,\ldots,q_N)$ are $N$-dimensional vectors. And $\chi_{[0,E]}(x)=1$ if $x \in [0,E]$ and $0$ otherwise. Unfortunately, I think you somehow need to have the right idea in order to calculate this integral and currently I am not seeing through this.
First, recognize that you're calculating the measure of a particular set. If we define $$f(p,q) = \sum_{i=1}^N {p_i^2 \over 2m} + {m\omega^2q_i^2\over 2} $$ then the set is $f^{-1}([0,E]) \subset \mathbb{R}^n \times \mathbb{R}^n$. Then rewriting $f$ by factoring out a ${1\over 2m}$ from the function and defining $r_i = m\omega q_i$, we find $$\int \chi_{[0,E]}(f(p,q)) dp dq = {1\over (m\omega)^n} \int \chi_{[0,2m\cdot E]}\left(\sum_{i=1}^N p_i^2 + r_i^2 \right)dpdr.$$ This right-hand integral is simply the volume of a sphere in $p-r$ space with radius $\sqrt{2mE}$, so $$\int_{\mathbb{R}^n \times \mathbb{R}^n} \chi_{[0,E]} \left( \sum_{i=1}^N {p_i^2 \over 2m } + {m\omega^2 q_i^2 \over 2} \right) dp dq = \left({2E \over \omega}\right)^n \textrm{Vol}\left( B_{2n} \right) $$ where $B_d$ is the unit ball in $\mathbb{R}^d$.