Given a bounded convex set $S\subseteq\mathbb{R}^3$ of unit width (not constant unit width, just that the minimal distance between two parallel planes enclosing $S$ is $1$), the volume $V$ can be as large as desired, by e.g. taking $S$ to be a $1 \times 1 \times X$ box for large $X$.
However, it is obvious at least intuitively that the volume is bounded from below, and it is not hard to obtain weak lower bounds. I can show that e.g. $V\ge \frac1{24}$, by measuring the width in three mutually orthogonal directions (I think such an approach can yield $V\ge\frac16$, but an elegant proof eludes me - the best I've been able to do so far is $\frac{5}{81}$).
The regular tetrahedron of side length $\frac{\sqrt{6}}{2}$ attains $V=\frac{\sqrt{3}}{8}\approx 0.2165$. Is this optimal? What are the best bounds we have on this problem?
In two dimensions, the minimal area among unit-width convex planar sets is $\frac{\sqrt{3}}3$, attained by the equilateral triangle. If the regular tetrahedron is indeed optimal in three dimensions, does the regular simplex attain the minimal volume in $n$ dimensions? Is there a known inequality akin to the isodiametric inequality for this problem in $\mathbb{R}^n$?