Your hypothesis that
if you have something like:
$$\frac{1}{kT}=\left(\frac{\partial S}{\partial E} \right)_{N,V}$$
And they've created a situation where entropy decreases with increasing energy.
is exactly right. The concept of negative absolute temperature, while initially counterintuitive, is well known. You can find a few other examples on Wikipedia.
In your question you say that temperature is "the average kinetic energy of ... particles". Strictly speaking, this is only true for an ideal gas, although it's often a good approximation in other systems, as long as the temperature isn't too low. It's slightly more accurate to say that temperature is equal to the average energy per degree of freedom in the system, but that's an approximation too - energy per degree of freedom would be $E/S$, whereas $T$ is actually proportional to $\partial E/\partial S$, as you say. It's much better to think of $\partial E/\partial S$ as the definition of temperature, and the "energy per degree of freedom" thing as an approximation that's useful in high-temperature situations, where the number of degrees of freedom doesn't depend very much on the energy.
As Christoph pointed out in a comment, the significance of the new result is that they have achieved negative temperature using motional degrees of freedom. You can read the full details in this arXiv pre-print of the original paper, which was published in Science.
You seem to make the implicit assumption that your vessel is placed in an environment that does not emit any thermal radiation, i.e. is already at 0 K temperature. The temperature of your container will asymptotically decrease to 0 K but will never actually reach it.
Assuming black-body radiation, fixed heat capacity $c$, and sufficient thermal conductivity, the temperature will decrease as $dT/dt = -A\sigma T^4/c$, with $\sigma=5.67\times10^{-8}~\mathrm{WK^{-4}m^{-2}}$ the Stefan-Boltzmann constant and $A$ the external area of the container. The temperature will decrease over time as $\propto t^{-1/3}$, which is rather slow.
Update
If you want to do the calculation more precisely, you'd first have to put a prefactor for the emissivity of your gas, as a function of temperature. Typically, gases at reasonable densities in man-like volumes have an emissivity very close to zero. Moreover, at some point, the gas would condense on the walls of your container (I think around 20 K for hydrogen and 4 K for helium, and much higher than that for anything else), which would turn your "gas in a container" into a much more difficult "solid state on a container wall" problem.
Only once you've figured out all that, you can start to wonder about more fancy quantum mechanics such as the probability that the entire crystal lattice makes the transition from its first excited state to the global ground state via a radiative transition. I'd wager that it will take more than the age of the universe to get there.
Update 2
Let's see how long it will take to approach the ground state. Suppose that the container has a size $L=1$ m; the lowest-energy transition is the one for a phonon with a wavelength equal to $L$. If the speed of sound in the condensed material is v=1 km/s, then the relevant temperature for the last transition to the ground state is $T_1=hv/(Lk_B)$=50 nK. If the emissivity of the gas is $\epsilon$, the time to reach the final temperature from an initial temperature $T_0$ is
$$ t = \frac{c}{3A\sigma\epsilon}\left(\frac{T_0}{T_1}\right)^3. $$
If I plug in some numbers, e.g. 1 kg of air (c=1 kJ/K), emissivity 1e-3 (wild guess), T1=50 nK, T0=300 K, A=1 m2, then I find t=1e+42 s, i.e., 1e+34 years. Yes, longer than the age of the universe before the quantum mechanics of the ground state start to play a role.
Best Answer
A super-thick sweater probably isn't the way to go - you may be better off wrapping yourself in aluminum foil.
The body loses heat through a handful of mechanisms:
(Note that your body emits radiation, but also receives it, with the amount depending on your particular radiation environment.)
Of these four mechanisms, the first two are irrelevant to your question because you are in a vacuum. Evaporation will definitely occur, especially around your nose, mouth, and eyes, but I think that the primary mode of heat loss here will be radiation, so let's focus on that.
Your body generates heat at all times via your metabolism as well as internal friction. If you are relaxing in comfortable conditions, you are producing roughly 100 W - but this number increases if you start to exercise. In particular, when your body gets cold your brain activates the shiver reflex, which can cause your body's power output to jump to 200-300 W.
Source (Note that $1 \text{ Cal/hr} \approx 1 \text{ W}$).
Ignoring for a moment the effect of clothing, then your equilibrium body temperature can be roughly estimated by equating the power generated by your metabolic processes (and possibly movement) with the power loss via radiation, assuming that you are not absorbing radiation from anywhere else. I am assuming that the body is at a uniform temperature here. This would not be the case - the core of your body would be warmest and then a gradient would form to your skin - but this can be neglected because the gradient would not be very extreme.
In this simplified model, this is the resulting equilibrium body temperature as a function of emissivity, assuming first 100 W and then 300 W of generated power.
As you can see, the situation is rather bleak if you're facing the void in the nude. Your core temperature can't drop much below its normal 37 C before you enter a hypothermic state; even shivering ferociously, this requires an emissivity of something like $0.425$, far below your body's typical value of $0.95$.
This is where clothing comes in. Textiles have a somewhat lower emissivity than naked humans do. The surface emissivity of wool is about 0.74, and most textiles are in that range or higher, which means that the surface of the garment would still equilibriate below 0 C.
However, the thermal conductivity of wool is only about $0.03\frac{\text{W}}{\text{m K}}$. For a garment of thickness $t$ covering your entire body, the temperature gradient from your body's surface to the surface of the garment would be
$$\frac{\Delta T}{t} = - \frac{100\text{ W}}{2\text{ m}^2 \cdot 0.03 \text{W/mK}} \approx 1670 \frac{\text{K}}{\text m}$$
Starting from the temperature of the garment's exterior, this allows us to track back and find the corresponding body temperature as a function of thickness. I've performed the calculation for wool and for cotton, with the results shown below.
The surface of a wool sweater would equilibriate at approximately -5 C, which would correspond to a 37 C body temperature if the thickness of the sweater were only about 3 cm. That's thick, certainly, but not absurdly so. For a cotton sweater, which would have both higher emissivity and higher thermal conductivity, the surface would equilibriate around -10 C and you would need a thickness of closer to 6 cm to keep you warm.
On the other hand, you could consider wrapping yourself in a layer of extremely low-emissivity material, and that would be much more effective. Polished silver, for instance, has an emissivity of only $0.02$, which would be problematic in the wrong direction. To radiate 100 W/m$^2$, our layer would need to have a surface temperature of about 60 C, which would roast us alive. The sweet spot - at which our body would equilibriate at 37 C - appears to correspond to an emissivity of approximately $0.15$. Based on this table of emissivities, it seems that alumel (an alloy of nickel, aluminum, manganese, and silicon) would do the trick.
Further Reading:
Convective and radiative heat transfer coefficients for individual human body segments
The Relative Influences of Radiation and Convection on the Temperature Regulation of the Clothed Body