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}