What is the kinetic energy of a quantum particle in forbidden region? I have read that if a particle is trapped in a finite potential well, it has a finite chance to tunnel out from it. Therefore, one can find a particle in a region where its potential energy is greater than its total energy.
I was wondering what happens to the kinetic energy of a particle in that region? 
What its value would be? 
Would it be negative and if so what does it physically mean?  
 A: The region in which the potential energy of a particle exceeds its total energy is called the "classically disallowed region".  In the classically disallowed region, mathematically, the term that represents the kinetic energy is, in fact, negative.  This changes the wavefunction from a "wave" that oscillates to a evanescent wave that decreases, typically exponentially.
I am not sure what you mean, otherwise by "what does this mean physically".  Possibly you mean "can this negative kinetic energy be measured, somehow"?  Or something similar about quantum measurement.  And, not quite knowing the question, I a not quite willing to figure out the answer to a question I might guess at.  Quantum measurement is quite paradoxical and hard to explain.
A: The kinetic energy is $p^2/2m$. To measure it means basically the same thing as to measure the momentum $p$, at least in the one-dimensional case and if we neglect the sign of $p$.
The uncertainty principle says that $x,p$ cannot be measured simultaneously and accurately.
To know that the particle is inside the wall (inaccessible region), we must first measure its position, with the precision better than $\Delta X$, the thickness of the wall. But once we do so, we unavoidably change the momentum $p$ of the particle, and therefore the kinetic energy, too. 
So if we measure the kinetic energy after the (precise or approximate) measurement of the position, we will get something else than what we would get before that – this is what $[x,p]\neq 0$ primarily means – and the energy conservation law doesn't hold if we naively substitute the values of $x,p$.
In other words, it's important to realize that in QM, the measurement of the total energy $T+V$ which depends on $x,p$ cannot be reduced to the measurements of $x$ and $p$ because those can't be done simultaneously. In QM, every observable (Hermitian operator) is measured by its own specific procedure. The accurate measurement of the energy must be done "at once", not by a decomposition to the measurement of mutually non-commuting observables.
When we become certain that the particle is located in a region/interval inside the wall, the wave function is projected so that it vanishes outside this interval. The wave function becomes a rather regular localized wave packet and its possible values of $p$ and $T$ are all non-negative. 
So the energy conservation law is "apparently" violated by the whole big amount, at least $V(x)-E_{\rm initial}$. However, the probability of this "arbitrarily large violation" goes to zero exponentially if this gap grows bigger. The correct interpretation isn't that the energy conservation law is actually violated – $H$ is exactly conserved.
Instead, the correct description is that if we try to measure the energy $T+V$ by measuring approximately $x,p$ after each other, we only get the approximate value of the whole energy $H$ and the probability that these two are "very different" is nonzero even for large differences, but it exponentially goes to zero (with the probability of the tunneling to the given place). The same conclusion would hold if you measured the kinetic energy first, and shortly afterwards, you would measure $x$ to find the particle inside the wall.
At any rate, one never measures any "imaginary momentum" or "negative kinetic energy".
