The Fourier transform $F(\mathbf{k})$ of the unit ball in L1 metric can be evaluated as follows$^\ast$.
Notation: $\theta(x)$ is the unit step function, $\delta(x)$ is the delta function, and ${\cal P}$ denotes the principal value of the integral.
$$F(k_1,\ldots k_n)=\int \cdots\int e^{i\mathbf{k\cdot x}}\theta\left(1-\sum_{p=1}^n|x_p|\right)dx_1\cdots dx_n$$
$$\qquad=\frac{1}{2\pi i}{\cal P}\int_{-\infty}^\infty \frac{ds}{s}\left(1-e^{-is}\right)\prod_{p=1}^n\left(\pi\delta(k_p-s)+\pi\delta(k_p+s)+\frac{2is}{s^2-k_p^2}\right)\qquad \mathbf{(1)}$$
$$\qquad =\frac{1}{2\pi i}{\cal P}\int_{-\infty}^\infty \frac{ds}{s}\left(1-e^{-is}\right)\prod_{p=1}^n\left(\frac{2is}{s^2-k_p^2}\right)$$
$$\qquad\qquad\qquad\qquad+\,\text{Re}\,\sum_{q=1}^n\left[\frac{1-e^{-ik_q}}{ik_q}\prod_{p=1,p\neq q }^n\left(\frac{2ik_q}{k_q^2-k_p^2}\right)\right].\qquad \mathbf{(2)}$$
Then finally I arrive at the result
$$F(k_1,k_2,\ldots k_n)=2\,\text{Re}\,\sum_{q=1}^n\left[\frac{1-e^{-ik_q}}{ik_q}\prod_{p=1,p\neq q }^n\left(\frac{2ik_q}{k_q^2-k_p^2}\right)\right].\qquad \mathbf{(3)}$$
I checked that for $n=2$ and $n=3$ this result agrees with the direct evaluation of the integrals (expressions given in the comment).
$^\ast$ Details of the calculation:
I first replace $\theta(1-\sum_p|x_p|)$ by $\theta(\xi-\sum_p|x_p|)$, denote the $\xi$-dependent integral by $F_\xi$, and take the derivative with respect to $\xi$, $F_\xi\mapsto F'_\xi$, so that the step function becomes the delta function $\delta(\xi-\sum_p|x_p|)$. Then I Fourier transform with respect to $\xi$, $F'_\xi\mapsto G_s$. The $n$-fold integral of $G_s$ over $x_1$, $x_2, \ldots x_n$ factorizes, so I can carry it out explicitly, using the identity $\int_0^\infty e^{ikx}\,dx=\pi\delta(k)+i{\cal P}k^{-1}$; finally I Fourier transform back $G_s$ to the $\xi$ variable, to obtain $F'_\xi$. One more integration of the $\xi$ variable, using $F_{\xi=0}=0$, and then setting $\xi=1$ gives the first integral expression (1).
For the second integral expression I may safely assume that all $k_p$'s are distinct, so products of delta functions do not contribute. I can then integrate out the single delta functions, obtaining the second integral expression (2).
The integral can then be performed by closing the contour in the lower half of the complex plane, the poles are shifted to the lower half of the complex plane, and the residue is reduced by a factor of two to account for the principal value. It then turns out that the integral equals exactly the contribution from the delta functions, so I end up with the final result (3).
