How to compute the determinant of this Toeplitz matrix?

The answer is: $f_n(x)=(1-x^2)^n$.

You can prove that this is true by induction. If you subtract from the first row the second row times $x$, all the entries of the first line after the first one become $0$ (and the first one is $1-x^2$). Therefore, $f_n(x)=(1-x^2)f_{n-1}(x)$. Since $f_1(x)=1-x^2$, you're done.

Subtract $x$ times row $2$ from row $1$, then $x$ times row $3$ from row $2$ etc. I get a lower triangular matrix with $n$ instances of $1-x^2$ on the diagonal and one $1$.

This is simply an illustration of José Carlos Santos' answer.

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Subtracting $x$ times the second column from the first gives $$ \begin{align} f_n(x) &=\det\begin{bmatrix} 1&x&x^2&x^3&\cdots&x^n\\ x&1&x&x^2&\cdots&x^{n-1}\\ x^2&x&1&x&\cdots&x^{n-2}\\ x^3&x^2&x&1&\cdots&x^{n-3}\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ x^n&x^{n-1}&x^{n-2}&x^{n-3}&\cdots&1 \end{bmatrix}\\ &=\det\begin{bmatrix} \color{#C00}{1-x^2}&x&x^2&x^3&\cdots&x^n\\ 0&\color{#090}{1}&\color{#090}{x}&\color{#090}{x^2}&\color{#090}{\cdots}&\color{#090}{x^{n-1}}\\ 0&\color{#090}{x}&\color{#090}{1}&\color{#090}{x}&\color{#090}{\cdots}&\color{#090}{x^{n-2}}\\ 0&\color{#090}{x^2}&\color{#090}{x}&\color{#090}{1}&\color{#090}{\cdots}&\color{#090}{x^{n-3}}\\ \vdots&\color{#090}{\vdots}&\color{#090}{\vdots}&\color{#090}{\vdots}&\color{#090}{\ddots}&\color{#090}{\vdots}\\ 0&\color{#090}{x^{n-1}}&\color{#090}{x^{n-2}}&\color{#090}{x^{n-3}}&\color{#090}{\cdots}&\color{#090}{1} \end{bmatrix}\\[6pt] &=\color{#C00}{\left(1-x^2\right)}\color{#090}{f_{n-1}(x)} \end{align} $$ Since $f_0(x)=1$, we have that $$ f_n(x)=\left(1-x^2\right)^n $$