How fast will sublimed dry ice mix with air? I saw this photo and wondered: Will the CO2 stay mostly in a layer on the floor with the rest of the atmosphere resting on top, or will it quickly diffuse throughout the room?

This lab is probably well ventilated, but assume it's not. What factors are involved in a back-of-the-envelope estimate to determine how quickly a given quantity of sublimed CO2 will diffuse throughout the room?
 A: The fog you are seeing is condensation of atmospheric water, not sublimed $CO_2$. The water fog is made very near the boiling surface, and then sinks slowly, exactly as it does in rainclouds. Therefore, just because you can see fog gathering on the floor does not mean that the $CO_2$ is confined there.
The $CO_2$ molecules have a speed, in random directions, of roughly $\sqrt{\frac{3\,k_B\,T}{m}}$. Even at the sublimation temperature ($195{\rm K}$), this works out to be of the order of hundreds of metres a second. Even accounting for the mean free path, this means that the gas in the room is going to mix very swiftly, notwithstanding the fog.
So you need to assume the $CO_2$ is spread evenly in the room for safety calculations. A kilogram of the stuff is about 23 moles, or about $\frac{23\,R\,T}{P}\approx \frac{1}{2}{\rm m^3}$ at room temperature. In a $4{\rm m}\times 2{\rm m}\times 2.54{\rm m}$ room, that amounts to about $2.5\%$ by volume. Although a kilogram brick dumped into water does take some time to sublime, I'd be doubling this as a safety margin and saying you definitely need strong, non recirculating fan ventilation or open windows in a small, office sized room. 
The short answer therefore is once you get to the sublimation of the order of hundreds of grams in an office sized room, you need to think about ventilation carefully. When I do this for children at my daughter's school (after we build and test a cloud chamber), I always open the windows.

Some More Quantification
The following is in response to user John Rennie's comment

From personal experience playing with various gases your estimate of the mixing rate is wildly optimistic. In the absence of convective mixing the mixing timescale is surprisingly long i.e. tens of minutes. 

Let's, out of interest, have a look at this more closely. It will NOT bear on a safety calculation. From the calculation of the mean free path derived in Wikipedia under the "Mean Free Path" heading, we get:
$$\ell = \frac{k_{\rm B}T}{\sqrt 2 \pi d^2 p}$$
which is of the order of, say $10^{-7}$ metres. The frequency of collision is therefore of the order of $\ell/v \approx 300/10^{-7})\approx 3{\rm GHz}$.
If we think of one Cartesian component of the motion, the sign of that motion is randomised with the collisions. Therefore, by the central limit theorem (if we make the resonable assumption that the signs are uncorrelated random variables), the displacement of the particle after $N$ collisions is a Gaussian random variable with a mean of nought but with a variance of the order of $N\, \ell^2/4$. When $\sigma=1{\rm m} = \sqrt{N}\ell/2$, we have $N\approx (2/10^{-7})^2\approx 4\times 10^{14}$. At a $3{\rm GHz}$ collision frequency, this amounts to a few hours. 
One should not be relying on this for safety calculations, so I don't think the above has any bearing on the spirit of my answer. 
