Binary black hole merger viewed from inside the event horizon How did the metric evolve inside the event horizons of the black holes whose merger caused the GW150914 signal?
In principle the Schwarzschild metric of a non-rotating black hole is known inside the event horizon, although the analogous Kerr solution for rotating black holes seems to have unphysical properties in this region. Is it possible to at least simulate the dynamics of the metric inside the event horizon during a black hole merger, and get a meaningful answer? If so, what happens and what would an observer inside the event horizon see? If not, why not?
The main inspiration for the question is my semi-Newtonian intuition that once the event horizons merge, the two singularities would rapidly orbit each other inside the event horizon, and eventually crash into each other due to emission of gravitational waves (which of course must remain trapped inside the event horizon). I highly doubt that this intuition is correct. Can general relativity give us a better answer?
 A: This might not literally be what you are aiming for, but everything we see or hear is from events before the horizon forms. And the gravitational waves are also emitted before the horizon forms.
Which means there are events in those stars and between those stars that happen after the waves we've seen are emitted. All the way in the center of the star we called the first black hole, and all the way in the center of the star we call the second black hole and in any part of the space in between.
So we do know what that looks like. And we can solve for it. And you do see the two stars orbiting each other. No singularities. No event horizons. Just time dilated signals coming out towards us.
So its "inside" in the sense that there are no holes or missing parts. It's just that the pre horizon events are very time dilated so a little bit of them covers a lot of our time. And at least those are the most scientific parts.
A: I was just thinking about this particular question, and I think I came up with an answer you probably won't like: it doesn't matter. Unfortunately for us (or fortunately?), anything that happens inside an event horizon is completely unknowable to the outside. While we can have a picture of the metric inside the event horizon for some distance, we should expect that the classical approximation breaks down the further we get. The fact that there is a singularity at the center of the black hole in GR indicates not that there is a region of infinite density but that GR as a whole breaks down on that scale.
Also, note that the gravitational waves emitted in the recent merger that we detected are not emitted from inside the event horizon, but from outside of the horizon - otherwise, they wouldn't escape, as you mentioned.
However, one paper I read described the shape of the photon sphere of the merger of (maximally charged) black holes, which has an analytic solution. I would expect (but I haven't done the calculation, so I'm not quite certain) that the event horizon takes on a form similar to that of the photon sphere (but smaller) during the actual merger event. Once the event horizon becomes that of a sphere, what happens on the interior is no longer relevant - everything outside the event horizon cannot tell the difference between a black hole with two singularities or a black hole with one singularity, so for all intents and purposes, we can consider it as having a single singularity.
One question that is possibly worth investigating, however, is exactly how the event horizons combine and the dynamics there - what happens when they overlap, or can they even overlap? Further, an interesting continuation of your question is, if you have two black holes (of equal mass) about to collide with impact parameter zero, and place a point particle exactly between the two bodies, what happens to that point particle? Clearly, the black holes do collide - however, from a distant observer's perspective, the particle can never pass the event horizon of either black hole. So does it get squeezed out? Spread out equally over both horizons? That's more to consider.
A: 
Binary black hole merger viewed from inside the event horizon

At the event horizon the "coordinate" speed of light according to distant observers is zero. So if you were viewing the merger from inside the event horizon, and I was somehow viewing you from where I'm sitting via my bubble of artistic licence, I would say you have no view at all. There's a conflict between this and "the proper speed is c". See Kevin Brown's The Formation and Growth of Black Holes and note the mention of future infinity. As far as I can tell it takes you forever to see anything. You haven't seen it yet, and you never ever will. 

How did the metric evolve inside the event horizons of the black holes whose merger caused the GW150914 signal? 

I don't know. But the black hole is a black hole for a reason, and it isn't because space is falling down. It's because the "coordinate" speed of light is zero. Which suggests there's no evolution inside the event horizon.   

In principle the Schwarzschild metric of a non-rotating black hole is known inside the event horizon

I rather thought that the Schwarzschild metric was only valid outside the event horizon, as per Graham Reid's comment.   

although the analogous Kerr solution for rotating black holes seems to have unphysical properties in this region. 

Agreed. Take a look at the Einstein digital papers, and you can read this from 1920: "the curvature of light rays occurs only in spaces where the speed of light is spatially variable". He didn't use the word coordinate, which suggests that at the event horizon, the speed of light is zero. Which suggests that the spin rate is zero too. 

Is it possible to at least simulate the dynamics of the metric inside the event horizon during a black hole merger, and get a meaningful answer? 

Maybe. But my reading of the Einstein digital papers is that there are no dynamics. Rather counterintuitively, the descending photon slows down. See this PhysicsFAQ article by editor Don Koks.  

If so, what happens and what would an observer inside the event horizon see?

Maybe there's a way we can have our cake and eat it, in that the expansion of the universe is not limited to the speed of light. But I'm still not hopeful about that observer seeing anything any time soon. 

If not, why not? 

Because the light doesn't get out. Because gravitational time dilation is infinite. Because the coordinate speed of light is zero. 

The main inspiration for the question is my semi-Newtonian intuition that once the event horizons merge, the two singularities would rapidly orbit each other inside the event horizon, and eventually crash into each other due to emission of gravitational waves (which of course must remain trapped inside the event horizon). I highly doubt that this intuition is correct. Can general relativity give us a better answer?

I think it can, but I side with the frozen-star interpretation, and I'm currently in the minority. Again see Kevin Brown's The Formation and Growth of Black Holes: "Incidentally, we should perhaps qualify our dismissal of the "frozen star" interpretation, because it does (arguably) give a serviceable account of phenomena outside the event horizon, at least for an eternal static configuration. Historically the two most common conceptual models for general relativity have been the "geometric interpretation" (as originally conceived by Einstein) and the "field interpretation" (patterned after the quantum field theories of the other fundamental interactions). These two views are operationally equivalent outside event horizons, but they tend to lead to different conceptions of the limit of gravitational collapse. According to the field interpretation, a clock runs increasingly slowly as it approaches the event horizon (due to the strength of the field), and the natural "limit" of this process is that the clock asymptotically approaches "full stop" (i.e., running at a rate of zero). It continues to exist for the rest of time, but it's 'frozen'..."
What this frozen-star interpretation says is that there aren't any point-singularities. Kevin Brown suggests Einstein would have sided with the other more common interpretation, but I don't think he would. When you drop your pencil, it falls down because the speed of light at the floor is less than the speed of light in front of your face. But when you're at the event horizon, the speed of light in front of your face is zero. And it can't go lower than that. 
A: A perfect symmetric view without new extensions : after the horizon, there is a privilegied direction. Each BH falls in the other. Nothing escape but the gravitational waves emited in relation with the shape.
From the point of view or each, all happens as if there were a star and a black hole.
