The dominant contributions for large $t$ come from the two regions around $\mathbf k_0 = \{0,\ldots,0\}$ and $\mathbf k_\pi = \{\pi,\ldots,\pi\}$, where $f(\mathbf k)\ll 1$. Note that $f(\mathbf k_0)=f(\mathbf k_\pi)=0$ are the only zeroes of $f$ in the integration interval. $ \DeclareMathOperator{\sign}{sign} \DeclareMathOperator{\sinc}{sinc} $
We first focus on the region around $\mathbf k_0$. Substituting $\mathbf k \mapsto \mathbf q/t$ and expanding the integrand around $t=\infty$ we find \begin{align} \frac{\sin(t f(\mathbf q/t))}{t\sin(f(\mathbf q/t))} = \sinc\left(\frac{|\mathbf q|}{\sqrt d}\right) + \mathcal O(t^{-2}) , \end{align} with the cardinal sine, $$ \sinc(x)=\frac{\sin x}{x}.$$ For the one-dimensional case $d=1$ this gives \begin{align} I^{(d=1)}_{0}(\infty;v)&=\int_{-\infty}^{\infty}dq \sinc(q) \, e^{i q v} \\ &=\frac{\pi}{2}[\sign(1-v)+\sign(1+v)], \end{align} which is the well known Fourier transform of $\sinc(q)$. The region around $\mathbf k_\pi$ contributes the same, but with an additional phase factor, such that in total \begin{align} I^{(d=1)}(\infty;v)&=\frac{\pi}{2}[\sign(1-v)+\sign(1+v)][1+\cos(\pi v)]. \end{align} In higher dimensions the calculation is quite similar, \begin{align} I^{(d)}_{0}(\infty;\mathbf v)&=\int_{-\infty}^{\infty}d^d \mathbf q \sinc\left(\frac{|\mathbf q|}{\sqrt d}\right) \, e^{i \mathbf q \cdot \mathbf v}, \end{align} which now is the $d$-dimensional Fourier transform of the cardinal sine, with argument $|\mathbf q|/\sqrt{d}$. In $d=2$, Mathematica can evaluate this transform, with result \begin{align} I^{(d=2)}_{0}(\infty;\mathbf v)&= \Re\left(\frac{4\pi}{\sqrt{1-2 |\mathbf v|^2}}\right). \end{align} The region around $\mathbf k_\pi$ should again contribute a phase shift. Note that the result for $I^{(d)}_{0}(\infty;\mathbf v)$ must be a function of $|\mathbf v|$, as the Fourier transform of an isotropic function is isotropic, see the update below. Maybe it can be generalized to general $d$, the op should check the literature. I must go to bed now...
Update #1
See, e.g., MO:315613 and references therein for a discussion of the case $d>2$.
In $d=3$, the solution seems to be (maybe up to a constant) \begin{align} I^{(d=3)}_{0}(\infty;\mathbf v)&= \sqrt{\frac{\pi}{2}}\frac{\delta(1-\sqrt3|\mathbf{v}|)}{\sqrt3|\mathbf{v}|}. \end{align}
