Yes, of course. First, the radius of the visible Universe is 46 billion light years. Note that it's more than 13.7 billion light years because while the light from the most distant places of the Universe was travelling here for 13.7 billion light years but all "meters" in the light's trajectory continued to expand so that the current radius is larger, 46 billion light years, where "current" means "measured on the slice of spacetime where all observers are 13.7 billion light years separated from the Big Bang by proper time".
However, cosmological measurements such as WMAP have measured the density to be very close to the critical density so the curvature to be something like $1.02\pm 0.02$. So if we assume that the topology is a sphere, its own curvature radius has to be about 50 times larger than the radius of the visible universe - or, approximately, the characteristic curvature radius of spacetime as measured now.
I don't know the exact figure - which sensitively depends on the error margin you assign to the measured total $\Omega$ - but it's of order 1 trillion light years.
This lower bound is of course much higher than what you can get if you allow the universe to be e.g. a flat 3-torus. In that case, you have to rule out double images etc. and the lower bound will be comparable to the radius of the visible Universe.
Also, note that the lower bound on the 3-sphere radius doesn't mean that the minimum volume has to be the volume of the 3-sphere even if the 3-sphere the correct local shape. There are other topologies that are locally 3-sphere, such as the "lens spaces" - the quotient of the 3-sphere with respect to the discrete $ADE$ groups such as the symmetry group of an icosahedron (translated from a subgroup of $SO(3)$ to a subgroup of $SU(2)$ which is embedded to $SO(4)$ so that it can naturally act on a 3-sphere). In those ADE cases, the total volume would be divided by the order of the group.