Converting Seconds to Millimeters The concept is that time is another dimension, complementary to those we can observe and measure directly.  For those three, I can take a ruler and measure how many millimeters one point in space is far from another.  If time is another dimension, and if I had the ability to observe that dimension as I observe the other three, I should be able to take my ruler and measure a distance in millimeters along that dimension as well.
This leads me to ask the question:  How many millimeters are in a second?  As in, if I take my ruler and measure between a point located in a 3D space at a particular time, measure along the time axis of spacetime to the same point in 3D space one second later (or earlier), I should be able to come up with a conversion ratio between millimeters and seconds, right?
If this is so, would there be a meaning to a question like How fast are we moving through spacetime, along the time dimension?  As in, if we are following a vector in spacetime, what is the meaning of the magnitude of the time component of that vector?
 A: Putting aside the fact that time and space are different, and that while they are unified as part of a single Lorentzian manifold there are important differences between "timelike" directions and "spacelike" directions, the conversion factor you're looking for is the speed $c$.
More specifically, there are $\delta X =  3\times 10^8 \frac{\mathrm m}{\mathrm s} \times 1 \ \mathrm{s} \approx 3 \times 10^{11} \ \mathrm{mm}$ in one second, insofar as light will travel $3\times 10^{11} \ \mathrm{mm}$ in one second (assuming flat spacetime, etc).

If this is so, would there be a meaning to a question like How fast are we moving through spacetime, along the time dimension?

For a particle moving with 4-velocity $\mathbf u$, the quantity $u^0 = \frac{d(ct)}{d\tau}$ is the number of ticks of my laboratory clock for every tick of the particle's wristwatch.  Its magnitude is always greater than $c$, and is a measure of the time dilation effect.  More precisely, $u^0/c$ is the factor by which time appears to be dilated for the moving particle.
A: Special relativity is derived from the observation that the speed of light $c$ is the same in all inertial frames and doesn't depend on the direction.  This means that in a time interval $\mathrm{d}t$ light will always travel a distance $c\,\mathrm{d}t$.  The distance between two points is $\sqrt{\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2}$.  Putting both together we have after a simple re-arrangement
$$ c^2\mathrm{d}t^2 - \mathrm{d}x^2 - \mathrm{d}y^2 - \mathrm{d}z^2 = 0.$$
The Poincaré group is the group of transformations of time and coordinates $(t,x,y,z)\to(t',x',y',z')$ that leave this quantity invariant.  In simpler terms, the Poincaré group is a way to systematically enumerate all coordinate frames that maintain this condition.  The probably more well-known Lorentz transformations are the special transformations in this group that correspond to coordinate frames which move at constant speed relative to each other.
So without going too deep into the weeds, the speed of light $c = 30\, \textrm{cm}/\textrm{ns}$ is what links time and space in special relativity, and thus $1\,\textrm{ns}$ and $30\,\textrm{cm}$ correspond to each other in the way that I think you had imagined when you asked this question.
