To calculate this, we need to know what tipping moment your table will suffer. This is pretty simple.
With the arm totally extended, it is $31''$ long (half of $62''$). You seem to be clamping the arm to the table, so I'm going to assume that the distance from the screen at the tip of the arm to the center of the table is equal to $31 + \frac{16}{2} = 39''$. The pivot point when tipping, however, will be one of the wheels, so we can subtract the distance from the post to one of the wheels. This isn't given, so I'm going to assume that distance is around $5''$, which gives us a total arm of $34''$.
So, now we have to calculate the tipping moment, considering the weight of the screen and of the arm.
$$M = 30\cdot34 + 17.5\cdot\left(3 + \dfrac{31}{2}\right) = 1344\text{ lb-in}$$
Now, this is already resisted by the weight of the post table (120 lb1), which can be considered concentrated at the post, generating therefore a moment equal to
$$M = 120 \cdot 5 = 600\text{ lb-in}$$
You therefore have here an excess of $1344-600 = 744\text{ lb-in}$ which you'll need to counter with additional weights. We can find the necessary weight by distributing it uniformly on the base, which gives us:
$$W = \dfrac{744}{5} = 148.8\text{ lb}$$
This is however only the nominal weight that's necessary. I would put in quite a bit extra weight to be sure the structure will hold in unexpected situations (someone bumping against the screen or some such).
The table states it can support 300 lb, so this is still acceptable:
$$148.8 + 17.5 + 30 = 196.3 < 300\text{ lb}$$
Hell, I personally don't see why not simply put 250 lb of weight, just as a safety precaution.
You also need to check the clamping of the arm to the table. For this, however, we need more information regarding the table (thickness, material) and of the clamp.
1 This is actually the ship weight, which means the table will be lighter. Get the correct weight and repeat these calculations. Make sure to also check the actual distance from the post to the wheels, which I assumed here at $5''$.