The assumption made, which is not at all exact, is that the velocity component of the onrushing air which is perpendicular to the surface of the spoiler is reduced to zero while the parallell component remains untouched. This is far from correct, so you have to remember we are talking about a ballpark figure.
If the velocity is V and the angle of the spoiler is a (measured from the velocity of the aircraft, the perpendicular velocity will be V*sin(a).
The dynamic pressure q = 1/2*rho*(V*sin(a))^2
The force on the spoiler, perpendicular to the surface: F = q*A
The retarding component of the force: Fr = F*sin(a)
I reiterate that this is far from scientific but will give you an idea about the magnitude of the forces involved.
Assuming M.85@FL330 ISA, V will be roughly 250 m/s and rho 0.4 kg/m^3. Assume a to be 70 degrees.
q = 11e3 N/m^2.
With a weight of, say, 100 metric tons, this means you get
100/11*sin(70) m/s of deceleration per square meter of spoiler, a little less than 1G.
Hint: Look into the change of momentum of the air deflected. That would probably be a more correct approach!
P.S. Way too late to be doing calculus... so be kind if I messed something up.
I thought I was doing good trying to avoid those airport hotels... and look at me now.