could this be something to do with the amount of pixels on the 7D sensor, it must be very dense to get that many on a crop sensor
Yes and no.
(physically) Directly no, the pixel size would have no bearing on the fact that there might be a residual image.
(electrically) However, due to the fact that the pixel size is smaller, there will be less light hitting each sensor. This would mean that each sensor would have a smaller range, from 0 to full to be considered at each 'time step' to determine what light has hit the sensor (bit confused that sentance).
Thus, this signal would be 'amplified' if necessary to make it easier to read in with the processor. Should there be any residual 'data' (charge) in the sensor, then this would be amplified too. The amount of residual (ratio) compared to the amount of charge required for a full signal will be increased (higher noise to signal) due to the fact that the range is decreased.
If you look at an item discharging, when it is full, the rate of discharge is high, when it has almost fully discharged, the rate of discharge is low. For a given technology, if, say, you loose 90% of your charge per second (per time-base), then you will have 10% left after 1, 1% after 2, 0.1% after 3 (roughly), at which point it could be considered all-but empty. If you use the same actual rate of loss (not the percentage), but have the maximum set at where 10% might have been, then you could consider the item discharged after 2 time-bases. However, if you then amplify this data to get to the 100% values, then you would have a real amount of charge left.
But that is kinda suposition based on what they are probably doing within their sensors (only a Canon eningeer involved with the original tech could say for certain)