Can the energy of a black hole's quantum field escape its event horizon? From what I've read, Hawking radiation is produced far outside the event horizon. The radiation is produced by the quantum field of the black hole outside the event horizon. As more of these radiations are emitted, the quantum field's energy decreases, hence the black hole's mass decreases.
My problem with this explanation is, it appears that the energy of the quantum field inside the event horizon could somehow ‘flow’ out of it, to fill the lost energy of the quantum field outside. Even if the energy in the quantum field inside the event horizon couldn't ‘move’ to the quantum field outside, wouldn't the energy of quantum field inside the event horizon stays same, while the outside of event horizon will be devoid of energy?
If such assumptions are wrong, how can Hawking radiation reduce a black hole's mass, without resorting to virtual particles or negative energy?
 A: Hawking radiation depends on a flow of negative energy. There is a negative energy density in the surroundings of the black hole. By analyzing the energy flux at infinity, one notices that the spacetime is losing energy as time goes by. However, no energy can escape the black hole, as you noticed, so what happens is precisely that you get a flux of negative energy flowing into the black hole. Of course, there is no need to resort to "particles", since everything can be defined in terms of fields.
To understand how this is physically viable, let me quote Hawking's seminal paper:

The above discussion shows that the particle creation is really a global process and is not localised in the collapse: an observer falling through the event horizon would not see and infinite number of particles coming out from the collapsing body. Because it is a non-local process, it is probably not reasonable to expect to be able to form a local energy-momentum tensor to describe the back-reaction of the particle creation on the metric. Rather, the negative energy density needed to account for the decrease in the area of the horizon, should be thought of as arising from the indeterminacy of order of $M^{-4}$ of the local energy density at the horizon. Equivalently, one can think of the area decrease as resulting from the fact that quantum fluctuations of the metric will cause the position and the very concept of event horizon to be somewhat indeterminate.

Also,

A renormalised [energy-momentum] operator which was regular at the horizon would have to violate the weak energy condition by having negative energy density. This negative energy density is not observable locally.

Notice that, in a quantum setting, negative energy densities are a prediction, not an issue. For an example, see, e.g., Sec. 1.2 of 1208.5399 [gr-qc] and/or Problem 6 of Chapter 14 of Wald's General Relativity. In short, if $\hat{T}_{ab}$ denotes the normal-ordered Klein–Gordon energy-momentum tensor and $t^a$ is some timelike vector, one can find a state $|\psi\rangle$ such that $\langle \psi | \hat{T}_{ab} | \psi \rangle t^a t^b < 0$ in a region of spacetime. Hence, one can have negative energy densities as a consequence of quantum effects.
I should point out that Hawking also writes

Thus it will give positive energy flux out across the event horizon or, equivalently, a negative energy flux in across the event horizon.

So in principle, both descriptions are somewhat equivalent. Of course, positive energy flowing out of the black hole is a problematic interpretation since it is acausal, while there is nothing wrong with negative energy flowing in, since quantum theory allows for negative energy densities.
A: The answer to your question is the gravitational field of the black hole, which already exists before the (for example neutron star) collapses into a black hole and, moreover, this gravitational field extends to infinity (so outside the horizon).
Since the gravitational field already exists for the collapsing star, when the collapse "happens", the event horizon (which is not a physical thing, rather, a boundary) happens to be so that the gravitational field extends outside of it too.

A black hole, however, can have an electric charge, which means there is an electric field around it. This is not a paradox because a static electric field is different from electromagnetic radiation. Similarly, a black hole has a mass, so it has a gravitational field around it. This is not a paradox either because a gravitational field is different from gravitational radiation.

How does gravity escape a black hole?
The fact that the gravitational field extends outside the horizon, is no contradiction. Nothing, no information or particles need to travel from inside, because the gravitational field of the collapsing star (for example a neutron star) already existed outside the boundary (that we later call event horizon), and so the energy of this gravitational field can (and in your example does) influence the quantum fields outside the horizon.

This is what is used in the Hawking radiation arguments, where the virtual loop at the event horizon has one particle interacting with the gravitational field and energy is supplied by it so that the other becomes real and exits as real.

Can virtual particles be 'boosted' into becoming real particles by fields other than gravity?
Now there are many interpretations of Hawking radiation, in this one, the extreme energy of the gravitational field gives this "boost" to the quantum fields, helping fluctuations to manifest as (seen from a far away observer) as photons coming from the (outside region of the) black hole. This radiation is interpreted as Hawking radiation, and the reason it decreases the mass of the black hole is because it takes energy from the gravitational field of the black hole. No need for particles to go in or out the horizon or anything to be influenced from inside the horizon (of course there are many equivalently good interpretations, including particles going in and out but these all originate from outside the horizon), but just energy taken from the gravitational field of the black hole, decreasing its mass, which corresponds to the fact that in mainstream physics, the mass of the black hole is derived from its gravitational field's energy (ADM, Komar, Bondi).
So again, the answer is the gravitational field (and its energy) of the black hole which already exists outside the horizon. Please note that there is a way to view this intuitively (if I understand correctly that you are assuming an external view of the black hole) where all the energy of the black hole is outside the horizon (because it never crosses in a fininte amount of time), and it all evaporates before the black hole could fully form for a far away observer.
A: When stuff falls in through the horizon it gets entangled with the virtual particles (quantum bubbles, disconnected closed one particle propagators) in the vacuum. The entanglement last long enough for the virtual particles to get promoted (by the strong curvature of spacetime) to real positive and negative energy particles. From the perspective of the infalling matter it takes the blink of an eye, while for an outside observer it takes long years.
The infalling negative energy particles reduce the mass, while the outgoing positive ones carry inside information because of the entanglement. Particles and antiparticles annihilate and radiate away as Hawking radiation. Virtual photon bubbles get promoted too into positive and negative energy parts and radiate directly outward and inward.
