Joe Marshall posted the details of his replication of the Cavendish experiment on his blog. At the bottom he didn't do the calculations, so I got excited and worked them out.

Here's my math:

Laws

F = ma (Newton's motion)
x = sin t/(2pi*T) (simple harmonic motion with period T)
F_s = -kx (Hooke's law)
F_g = G m1 m2 / rad^2 (newton's grav.)

Observations

m1 = 0.9kg
m2 = 3.6kg
T (period of oscillation) = 40 min

To find the distance between centers of mass, for the G calculation, I had to guess based on the objects he used.

He said 1 inch, I assumed this was one inch separation between edge of the water jug and edge of the lead cylinder. Assumed radius of milk jug = 9cm; lead cylinder = 2.5 cm; separation = 2.5cm, for a total of 14cm between center of mass. 

rad_g = 0.14m

I also had to assume he placed the centers of the lead cylinders at the end of his 3ft dowel. 

rad_dowel = 1.5ft

And I also assumed that he was projecting the laser onto a perpendicular wall (20ft away): 

rad_wall = 20ft

light moved "1 or 2 inches": I assumed 3cm 

delta_x_wall = 0.03m

Calculations

First, find the spring constant, k, for the hanging system.

Doubly differentiating the harmonic motion equation and solving leads to: 

a = -x / (2pi * T)^2

Substitute f = kx and f = ma 

k = m / (2pi * T)^2

Use m = 2m1 (2 lead weights in motion). Neglect the mass of the dowel. I got 

k = 1.9 * 10^-5 kg/sec^2

Now we can figure out G. By similarity of triangles, 

delta_x = delta_x_wall * rad_dowel / rad_wall

By hooke's law, F_spring = k (delta_x). Since we are concerned with the system's new rest position, total forces are zero so F_gravity = F_spring.

There are two instances of gravitational interactions we're measuring which sum: 

F_spring = F_gravity
k*(delta_x) = 2 * G * m1 * m2 / rad_g^2

Plug in values and solve for G. 

G = 1.32 * 10^-10

whereas Google says 

G = 6.67300 * 10^-11 m^3 kg^-1 s^-2

(too big by about a factor of 2)