How much could one slow down aging using existing spacecraft? Imagine that I board a spacecraft and head away from the Earth at top speed until I've aged twenty years, then I turn around and come back. How much younger will I be than the unexciting folk who hung around Earth the whole time?
I realize that the practical impossibility of doing this could complicate answering the question: how do I avoid running out of energy (or food or oxygen, if you prefer)? How much fuel will I need to pull a 180 in deep space? So let's just say that the spacecraft can accelerate (let's say instantly) until it's moving away from the Earth at the speed of the fastest currently-existing spacecraft that has been used to transport a human and can turn around instantaneously and head back at the same speed. 
 A: 
So let's just say that the spacecraft can accelerate until it's moving
  away from the Earth at the speed of the fastest currently-existing
  spacecraft

First, note that the fastest speed, relative to Earth, that a spacecraft has obtained is an exceedingly small fraction of the $c$ and, thus, one should not expect significant time dilation.
For example, the New Horizons spacecraft achieved a relative speed of over 10 miles-per-second (mps).
However, the speed of light is about 186,000 mps.  To get an idea of how little time dilation to expect, let's plug this into the time dilation formula:
$$\Delta \tau = \Delta t\sqrt{1 - \left(\frac{10}{186,000}\right)^2} = 0.999999998\Delta t$$
So, making the simplifying assumption that your speed, relative to Earth, is constant for 20 Earth years, your age according to Earth clocks would be
$$0.999999998 \cdot 20y = 19.9999998y = $$
In other words, according to Earth clocks, your clock would be less than 1 second behind.
Now, this isn't a full calculation but, rather, a hint of how to proceed and to not expect too much in the way of a slow down.
A: Almost none.
Let's be much more generous than your idea of human-carrying craft.  Let's just use the fastest probe.  The Helios II craft, after nearing the sun, reached a heliocentric speed somewhere near 70 km/s.  Obviously, its speed was more due to the gravitational influence of the sun than its engines.  
$$t = \frac{t_o}{\sqrt{1 - \frac{v^2}{c^2}}} $$
$$t = \frac{20 years}{\sqrt{1 - 5.45\times10^{-8}  }} $$
$$t = \frac{20 years}{0.999999973}$$
$$t = 20.000000545 years$$
That's about 17 extra seconds over the 20 years.  Double that for the inbound and outbound journeys. Real-world technology would allow for a fraction of that result.
A: From ScienceMuseum:

Apollo 10 holds the record as the fastest manned vehicle, reaching speeds of almost 40,000 km per hour (11.08 km/s or 24,791 mph to be exact) during its return to Earth on 26th May 1969.

Using the formula (as above). After traveling for 40 years, you would be a little over 0.86 seconds younger.
Added:
I did some calculating and determined that if your radial velocity directly away from the sun is 11.08 km/s, you would still be inside the solar system after 20 years (just beyond Neptune).
A: Instead of using existing spacecraft, let's use a photon rocket powered by the gamma rays emitted by matter anti-matter annihilations in its reacor.  Where does the anti-mater come from? We will produce it using solar energy. We'll use giant solar panels that generate a huge voltage in vacuum which leads to Swinger pair production. Moving away from the Earth and back involves 4 changes in the velocity so from a given initial mass $m_i$ and a final mass $m_f$ we will can travel at a gamma factor of:
$$\gamma = \frac{1}{2}\left[\left(\frac{m_{i}}{m_{f}}\right)^{\frac{1}{4}} + \left(\frac{m_{f}}{m_{i}}\right)^{\frac{1}{4}}\right]$$
Let's then measure time in a unit define as the time it takes for the factory to produce an amount of fuel of mass $m_f$ (half of this is then anti-matter). So, if we wait 100 units of time, we'll have the fuel for travelling at a gamma factor of 1.74.
Now, the reason why we want to do this is because we recieved a message from E.T. and we replied back asking about the TOE. E.T.'s answer is expected to come in at about 10.000 units of time in the future. The optimization problem is thus to minimize the proper time needed to move 1000 units of time into the future. Suppose we run the factory for time T, and then we use the fuel to travel and return when the message comes in (obviously we would intercept the signal sooner if we keep on moving in the direction from where the signal will be coming, but let's suppose that we can only work on the TOE from home). The proper time is then:
$$\tau(T) = T + 2 (10^4 - T)\left[\left(T+1\right)^{\frac{1}{4}} + \left(T+1\right)^{-\frac{1}{4}}\right]^{-1}$$
The optimum value is then to choose T = 1075 which yields a proper time of 4099.4. However, we can do a lot better by using the fuel produced after a shorter time $T_1$ and then make a round trip using that fuel and then travel again by using the fuel produced in that travel time. E.g. if according to the factory's clock we start to travel at time of 170, come back at a time of 1200 and then arrive back at 10,000 to read E.T's message, we'll only have aged 3636 units of time.
