Chubbs, the alligator in the question, is 15 feet long, according to the OP's link. OP's source and mine: Sports Illustrated. According to this source
The state's [Florida's] biggest alligator on record is 14 feet long and 780
The article does not say if the previous record holder (PRH) had recently ate or pooped, which might make a non-trivial difference, if the OP desires great accuracy and precision.
I'm glad the OP asked for an estimate, not a measurement, otherwise one would have to sedate the alligator and suspend him (most likely a him) in a sling attached to a spring balance.
For a zeroth order approximation for Chubbs, one would take (15/14) x 780 and get 836 pounds.
Of course Chubbs, if he had the exact morphology of the (PRH), would be larger in all three dimensions. This would give a first order approximation of (15/14)cubed x 780 = 959 pounds.
To refine that number, one would look at the largest diameter (or circumference) of both alligators and adjust accordingly. Maybe Chubbs (named after a local golfer, and not his figure) is svelte relative to the PRH -- or obese. You should be able to get a reasonable approximation to the ratio of diameters from photos, the photos normalized to a 15/14 length ratio. I don't advise using a measuring tape on an unsedated alligator.
Then the estimate of Chubbs's weight would be (15/14) x [D(Chubbs)/D(PRH)]squared x 780.
For any other alligator, use the same method with a reference alligator as close as possible in length and shape.