Why Monte Carlo integration is not affected by curse of dimensionality?

What is the common sense explanation behind that fact that MC integration is free of "curse of dimensionality" in contrast to deterministic integration rules (e.g. trapezoidal rule)?

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This is because the normalization $1/\sqrt{n}$ of the central limit theorem does not depend on the dimension. – Did Dec 4 '11 at 14:32

To see what goes wrong, let's look at the tensor product of two linear one dimensional functions. We will get a bilinear function, but this function won't be linear unless the coefficient of the $xy$ term is zero. In general, a $n$-linear function $f: \mathbb{R}^n\to\mathbb{R}$ will have $2^n$ degrees of freedom whereas a linear function $l: \mathbb{R}^n\to\mathbb{R}$ just has $n+1$ degrees of freedom. I should add that what I mean by "linear function" here is a polynomial of degree $1$ or $0$. The "correct" meaning of "linear function" would be something like $l(a+b)=l(a)+l(b)$ and $l(\alpha a)=\alpha l(a)$.