We can extract something like a rule of thumb from the equation for the power going into gravitational waves by a pair of isolated, not-very-close masses in circular orbits around their barycentre:

P = -32/5 G^4/c^5 \frac{ (m1 m2)^2 (m1 + m2) }{r^5}.

The key adjustable here is the orbital distance r. As it goes down, power goes up, but it's still extremely even for large masses until the two masses are practically in contact. You can plug in kilograms for masses m1 and m2 and use metres for r5. We are already doing a linearized approximation here, so we can bite the bullet and circularize the orbit of Earth-Moon at the average orbital distance and find a value that's a few microwatts; doing the same for Earth-Sun gives us a couple of hectowatts. Putting a pair of tens-of-solar masses at r ~ 1000 km gives us a much much larger power.

We can take the integral of power P over time and arrive at the radiated energy. We can in turn consider an r that shrinks over time. We can also flip this on its head and consider that it is equally reasonable to say that r shrinks in response to energy loss or energy loss increases in response to r shrinking. More on that below.

The calculation grows much more complicated for realistic orbits especially when the orbital distance is small, but you only asked for a rule of thumb, not a detailed explanation of, say, the Bondi-Sachs mass loss equation. :-)

> Do you know why [large mass] were converted into gravitational energy?

Sure. Very roughly, angular momentum depends on frame of reference. If you choose to work in a frame of reference in which a black hole binary has significant angular momentum in the mutual orbit (e.g. in the cosmological frame), then you have to get rid of a chunk of that for two reasons.

Firstly, angular momentum needs to drop in order for bodies of constant mass to drop into a closer mutual orbit. Although there are other interactions at work when the distances are large for a binary BH in a galactic core, when the binary BHs are near the end of their inspiral, the orbital distance shrinks very quickly. The only reasonable mechanism to shed that angular momentum that quickly is gravitational radiation.

Secondly, the "no hair" theorem means that there is only a three-component angular momentum J an at any given time coordinate for a black hole once it has stabilized. But as each of the merging BHs has its own nonzero angular momentum there are excess multipole moments that count as hair on the asymmetric configuration at the end of the inspiral, and so balding has to happen. This causes excess momentum either to be swallowed into the merged black hole (which will have some final angular momentum) or radiated to infinity. Swallowing is limited by the final mass of the merged black hole. Superextremal black holes are forbidden, for example, and in practice merged BHs are unlikely to rotate near he maximum possible speed for their mass. Consequently the rest must disappear as gravitational radiation, since it can't escape in any other way.

> could [the whole configuration be radiated away]

A real answer would be quite deep since mass in General Relativity is not so straightforward. However, the individual black hole masses m1 and m2 will change very very little during the inspiral and merger; and the end black hole will be very close to m1+m2. What gets radiated away is maybe best understood as some of the quantity that works against the BHs simply falling straight down onto one another at the speed of light, which is something no observer sees. I discussed this a bit more at https://news.ycombinator.com/item?id=15353970