I'd like to prove the following two results, but besides the "trivial" implications, I haven't been able to crack them:
$$ \int_{(0,1)^n} \frac{1}{x_1^{\alpha_1}+x_2^{\alpha_2}+...+x_n^{\alpha_n}} dm_n < +\infty \Leftrightarrow \sum_{i=1}^n \frac{1}{\alpha_i} > 1$$ $$ \int_{(1,\infty)^n} \frac{1}{x_1^{\alpha_1}+x_2^{\alpha_2}+...+x_n^{\alpha_n}} dm_n < +\infty \Leftrightarrow \sum_{i=1}^n \frac{1}{\alpha_i} < 1$$
I have supposed $\alpha_k>0$ for all $k$. For $x_1^{\alpha_1}+\cdots+x_n^{\alpha_n}>0$, you have:
$$\int_0^{\infty}\exp(-t(x_1^{\alpha_1}+\cdots+x_n^{\alpha_n}))dt=\frac{1}{x_1^{\alpha_1}+\cdots+x_n^{\alpha_n}}$$ Hence using Fubini
$$\int_{[0,1]^n}\frac{dm_n}{x_1^{\alpha_1}+\cdots+x_n^{\alpha_n}}=\int_{0}^{+\infty}\prod_{k=1}^n(\int_0^1\exp(-tx_k^{a_k})dx_k)dt$$
Now $$\int_0^1\exp(-tx^{\alpha})dx=\frac{1}{\alpha t^{1/\alpha}}\int_0^t\exp(-u)u^{1/\alpha-1}du$$ Remark: For $t>0$ we have $$\frac{1}{\alpha t^{1/\alpha}}\int_0^t\exp(-u)u^{1/\alpha-1}du \leq 1$$ And your integral is $$\frac{1}{a_1\cdots a_n}\int_0^{+\infty}\frac{\prod_{k=1}^n\int_0^t\exp(-x_k)x_k^{1/a_k-1}dx_k)}{t^{1/a_1+\cdots1/a_n}} dt$$
Now the convergence at $0$ is easy using the remark, and as the integrals $\displaystyle \int_0^{+\infty}\exp(-x_k)x_k^{1/a_k-1}dx_k$ are convergent, it easy to finish the proof of your first assertion. I suppose that there is a similar proof for the second assertion.