Let $f(x)$ be a smooth function from $\mathbb{R}$ to $\mathbb{R}$. Let $\mu$ be a compactly supported Borel measure (not necessarily positive) on $\mathbb{R}^n$. Define $$ \tilde{\mu}(\xi) = \int e^{i (\xi_1 f(x_1) + \ldots + \xi_n f(x_n))} \mu(dx). $$ I have estimates of the form $\tilde{\mu}(\xi) \leq C(\xi)$ for any $\xi \in \mathbb{C}^n$. My question is if it possible to find the similar estimates on Fourier transform $$ \hat{\mu}(\xi) = \int e^{i \xi x} \mu(dx) $$ using given estimates on $\tilde{\mu}(\xi)$?

As stated, no. Let $f \equiv 0$, and $\mu$ with $\int \mu = 0$. You then have $\tilde{\mu}(\xi) = 0$ for any $\xi$, and you have no control over the nonzero frequencies $\hat{\mu}$. 


If your $f$ is onetoone, make the change of the variable in your integral $y_i=f(x_i)$. You will obtain a new measure $\mu_1$, also with compact support, and your $\tilde\mu$ will be the ordinary Fourier transform of this new measure $\mu_1$. 

