Definition of event horizon – gravity around black holes Put simpler:
A black hole is surrounded by a 'sphere' where, to an outside observer, the speed of light is (near) zero.
What is the radius of that 'sphere' (for a non-rotating black hole) and how strong is gravity there?
Definition of Event Horizon (EH)
A black hole is defined as a body whose mass falls within its Schwarzschild
radius $r_\mathrm S$, where the escape velocity $v_\mathrm e$ equals the speed of light $c$.
That is clear, nothing can escape from it. But, with the Event Horizon (EH)
described as a boundary that cannot be crossed, EH does not coincide with $r_\mathrm S$.
At $r_\mathrm S$, and everywhere in a gravitational field, only the acceleration of
gravity $g$ is apparent. $v_\mathrm e$ is derived, the equivalence of endless
decelerations by $g$. A particle does not directly experience $v_\mathrm e$.
Some arithmetic with $v_\mathrm e=(2GM/r)^{1/2}$ and $g=GM/r^2$, shows, in round numbers:


*

*A 'small' black hole of 3.3 solar masses ($M_\mathrm s$) – $r_\mathrm S$ is $10\ \mathrm{km}$ and at $r_\mathrm S$,
$g$ is a crushing $4.5\times10^9\ \mathrm{km/s^2}$. Its sphere of no return, where
$g$ is $300\,000\ \mathrm{km/s^2}$ is further out at $1200\ \mathrm{km}$.

*For one comparable to our Milky Way's of $3.3\times10^6M_\mathrm s$, $r_\mathrm S=1.0\times10^7\ \mathrm{km}$ with
$g$ at $r_\mathrm S$ a puny $4500\ \mathrm{km/s^2}$. Its EH is closer in at $1.2\times10^6\ \mathrm{km}$.

*The only instance where $g$ equals $300\,000\ \mathrm{km/s^2}$ at $r_\mathrm S$ is for a black hole
of $50\,000M_\mathrm s$ with $r_\mathrm S$ at $1.5\times10^5\ \mathrm{km}$.


Clearly I'm missing something.
 A: It is a truly unfortunate coincidence that the "escape velocity" at the Schwarzschild radius $R_\mathrm{S}$ is $c$. This leads people to think Newtonian mechanics can predict black holes. It can't.
Note that a Newtonian escape velocity is the minimum velocity needed to coast to infinity against gravity. If you have less speed, you still travel away from the surface some amount before being pulled back down, whereas nothing is able to move any distance at all out of the event horizon. Also, one can always escape a Newtonian gravity source at any speed with a rocket, whereas no rocket can help escape the event horizon.
So what is the event horizon? It is the surface that separates parts of spacetime that can transmit a signal to infinity from those that cannot. Technically one has to know the details of all the gravity and matter and energy in all of spacetime to know the location of the event horizon (you might think you can transmit a signal, but what if you are inside a black hole and don't yet realize it?). We only have simple formulas in very simply hypothetical cases (e.g. a single black hole in an empty universe that has been sitting unchanging for all eternity).
Acceleration doesn't have much to do with event horizons. You will need to accelerate more to hover above a small black hole compared to a large one, but that's about it. Note that $c = 300{,}000\ \mathrm{km/s}$, but an acceleration of $300{,}000\ \mathrm{km/s^2}$ has no special significance, just as a distance of $300{,}000\ \mathrm{km}$ is not a special distance.
A: In general relativity, an event horizon is a boundary in spacetime beyond which events cannot affect an outside observer, i.e. any events separated by an event horizon are space-like. An other way to say this is that any worldline with a start within the event horizon will never cross the boundary of the event horizon.
If you are familiar with the light cone from special relativity, you could imagine all events in the future light cone as being inside the event horizon. A such light cone would have its origin at or inside the event horizon, and thus there's no way for an event to cause something outside the event horizon, because all such worldlines would be spacelike.
(Note that the light cone below depicts (flat) Minkowski space, and not curved spacetime)

A: What you're missing is the second paragraph here:

That's Einstein saying light curves because the speed of light is not constant$^*$. The force of gravity at some location is related to the gradient in the speed of light at that location. See Baez: "This difference in speeds is precisely that referred to above by ceiling and floor observers". Raise your pencil to the ceiling and drop it. When it hits the floor its speed is related to the difference in the speed of light at the ceiling and the speed of light at the floor. In similar vein the event horizon isn't really a place where escape velocity is 299,792,458 m/s. Instead it's a place where the difference between the speed of light here and the speed of light there is 299,792,458 m/s. Because the speed of light there is zero. That's why like Chris said, nothing is able to move any distance at all out of the event horizon. Including light. Hence as you can read here at the event horizon of a black hole the coordinate speed of light is zero. Sadly the article then confuses matters by saying the proper speed is c, but nevermind. Forget that and follow the link to the Shapiro delay where you can read this: the speed of a light wave depends on the strength of the gravitational potential along its path. Here's a screenshot of Shapiro's paper:

A black hole is a place where gravitational potential is such that the speed of light is zero. That's why the light can't get out. That's why the black hole is black. 
* We nowadays tend to call it the "coordinate" speed of light, but Einstein didn't. Nor did Shapiro. 
Edit: Can I politely add that Newtonian mechanics did predict black holes. We trace them back to John Michell in 1783. See the Wikipedia article: "If there should really exist in nature any bodies, whose density is not less than that of the sun, and whose diameters are more than 500 times the diameter of the sun, since their light could not arrive at us..." As to whether this was a correct prediction, check out Newton's Opticks query 20: "Doth not this aethereal medium in passing out of water, glass, crystal, and other compact and dense bodies in empty spaces, grow denser and denser by degrees, and by that means refract the rays of light not in a point, but by bending them gradually in curve lines?" It's surprisingly similar to Einstein's explanation of why light curves.    
