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I am learning Fourier transform and I was wondering what the Fourier transform of $f(a_1 x_1, ..., a_nx_n)$ is with respect to $\hat{f}$? I could only find answers for the one dimensional case and I was wondering how this worked for $n$-dimensional. Thanks!

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It scales pretty much how you'd imagine it would.

Suppose we have the Fourier transform of $f$, given by: $$ \mathscr{F}_f(x_1,\ldots,x_n) = \int\limits_{-\infty}^{\infty}\ldots\int\limits_{-\infty}^{\infty} f(t_1,\ldots,t_n) \exp\left(-2\pi i\sum\limits_j x_jt_j \right) \,dt_1 \ldots \,dt_n $$ Let the scaled function be $\widetilde{f}=f(a_1t_1,\ldots,a_nt_n)$. Then we get: \begin{align} \mathscr{F}_\widetilde{f}(x_1,\ldots,x_n) &= \int\limits_{-\infty}^{\infty}\ldots\int\limits_{-\infty}^{\infty} f(a_1t_1,\ldots,a_nt_n) \exp\left(-2\pi i\sum\limits_j x_jt_j \right) \,dt_1 \ldots \,dt_n \\ &= \int\limits_{-\infty}^{\infty}\ldots\int\limits_{-\infty}^{\infty} f(a_1t_1,\ldots,a_nt_n) \exp\left(-2\pi i\sum\limits_j \frac{x_j}{a_j} \underbrace{t_ja_j\displaystyle}_{v_j\displaystyle}\displaystyle \right)\displaystyle \,\frac{dv_1}{|a_1|} \ldots \,\frac{dv_n}{|a_n|} \\ &= \left[ \prod\limits_k \frac{1}{|a_k|} \right] \int\limits_{-\infty}^{\infty}\ldots\int\limits_{-\infty}^{\infty} f(v_1,\ldots,v_n) \exp\left(-2\pi i\sum\limits_j \frac{x_j}{a_j} v_j \right) \,dv_1 \ldots \,dv_n \\ &= \left[ \prod\limits_k \frac{1}{|a_k|} \right]\mathscr{F}_f\left(\frac{x_1}{a_1},\ldots,\frac{x_n}{a_n}\right) \end{align} which is the analogue of the 1D case. This can be considered a Similarity Theorem for a Fourier Transform.