Dear John, let me post the same thing that Marek has said as a standard answer.
Einstein's equations are not equations for the Riemann tensor because the Riemann tensor's components are not independent fields. Instead, Einstein's equations are differential equations for the metric tensor.
In 4 dimensions, the metric tensor has 10 components - a symmetric tensor - and Einstein's equations have 10 components - a symmetric tensor - too. It doesn't matter that the Riemann tensor has 20 components because these 20 functions of space and time are calculated from the 10 component functions of the metric tensor and its (first and second) derivatives.
In fact, the 10-10 counting is oversimplified. Four "differential combinations" of Einstein's equations vanish identically because $\nabla_\mu R^{\mu\nu}=0$ is an identity (that always holds, even if the equations of motion are not satisfied). The same identity holds for the corresponding other tensors that are added to the Ricci tensor in Einstein's equations.
So instead of 10 equations, the Riemann equations are, in some sense, just 6 independent equations. That means that they don't determine the metric completely: they leave 4 functions undetermined and these are exactly the 4 functions that you may choose arbitrarily to specify a diffeomorphism, mapping one solution into another (equivalent) solution.
Up to the coordinate transformations which are always allowed to be made, initial conditions for the metric and its first derivative determine the metric tensor - and therefore the whole Riemann tensor - everywhere in the future. One doesn't need any Newtonian equations as a "mandatory supplement" in general relativity. That doesn't mean that the Newtonian limit is unimportant: of course, it is one of the most important approximate consequences of general relativity.
1) I gather you mean gravitation potential energy of the test particle. Well, any such thing is only useful in so far as it is related to a constant of motion throughout the geodesic--in the case of gravitational potential, being part of the conserved mechanical energy, kinetic + potential. (Another example could be angular momentum.)
In GTR, these constants are given by a Killing vector field, which is an infinitesimal generator of an isometry: spacetime "looks the same" in the direction of a Killing vector. Most spacetimes do not have any, but by definition, a static spacetime has a timelike Killing vector field, and can always be put in the following form:
$$ds^2 = -\lambda dt^2 + d\Sigma^2,$$
where $d\Sigma^2$ is the metric for any spacelike manifold and $\lambda$ is independent of $t$. The factor $\lambda^{1/2}$ is commonly called the gravitational redshift.
For example, for the Scwarzschild spacetime in the usual Schwarzschild coordinates, $\lambda = \left(1-\frac{2GM}{c^2R}\right)$, and the following is a constant of motion representing the specific (per-mass) energy of the freefalling particle:
$$e = \left(1-\frac{2GM}{c^2r}\right)\frac{dt}{d\tau}.$$
This is the natural generalization of the total mechanical energy, including also rest-mass energy; for the Schwarzschild case, the spherical symmetry allows one to build an "effective potential" quite analogous to the Newtonian case, but that approach is less useful in general.
2) Gravitational energy cannot be explicitly included in the Einstein field equations because the equivalence principle--there is always a local inertial frame (the free-falling one) in which spacetime looks like the ordinary, flat, special-relativistic one. Hence if there was a frame-independent local notion of gravitational energy, i.e., a tensor, that tensor is zero in some local frame, and hence zero in every frame.
However, one can think of the non-linearity of the Einstein field equation as caused by gravitational energy itself interacting with spacetime. In this sense, gravitational energy is "implicitly" included. Another thing one can do is try to build another notion of gravitational energy that's not necessarily both local and frame-independent, e.g., Landau-Liftshitz pseudotensor and others.
Best Answer
UPDATE
This answer raised some controversy, which I believe comes from the fact I didn't properly define what I mean by energy. In this post, whenever I say energy I refer to the canonical/Hilbert definition of the energy-momentum tensor, i.e., the functional derivative of the Lagrangian with respect to the metric. In GR, this is the energy that matters, because it directly enters in the EFE.
Note that this definition, in principle, has nothing to do with the classical interpretation of energy(=work), which is conserved along the trajectories of point particles. In GR we don't care about this latter energy, because it is not conserved in general (you need some Killing fields to exist, but OP asked us not to mention Killing fields).
Finally, in some semi-classical analysis of GR we can define some kind of potential energy (e.g., $2\phi(r)\sim 1+g_{00}(r)$), which does behaves like the classical interpretation of energy(=work). But this semi-cassical analysis, where you mix Newtonian and GR concepts, can only be used for weak or highly symmetric fields.
With this in mind, note that you can have bodies accelerating/slowing down because of the action of gravity. This doesn't contradict the fact that gravitational energy doesn't exist: when we say energy is conserved in GR, we mean $\nabla T=0$, but this energy is the canonical/Hilbert energy, not the kinetic+potential energy of the bodies! If you want to talk about kinetic+potential energy of a test body, you'll need Killing sooner or later!
First Question
The Einstein Field Equations tell you how curvature is related to energy:
$$ R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=T_{\mu\nu} $$ (where I use natural units $c=8\pi G=1$)
The l.h.s. is a measure of curvature, while the r.h.s. is a measure of the energy of the system. Once you know the r.h.s, you can solve these PDEs to find the metric (l.h.s).
Some examples:
You probably know from classical electromagnetism that the electromagnetic field carries energy (something like $\boldsymbol E^2+\boldsymbol B^2$ should look familiar). In tensor notation, the energy is written $$ T^{\mu\nu}=F^{\mu\alpha}F^\nu_\alpha-\frac{1}{4}g^{\mu\nu}F^2 $$ where $F$ is the electromagnetic strenth tensor (note that $T^{00}\propto \boldsymbol E^2+\boldsymbol B^2$).
If you have some matter around that you can model as a perfect fluid (e.g, negligible viscosity), then the formula for the energy is given by $$ T^{\mu\nu}=(\rho+p)u^\mu u^\nu+p g^{\mu\nu} $$ where $\rho,p$ are the mass density and the pressure of the fluid, and $u$ its velocity.
Etc
Anything that is present in the system generates gravity, and thus you should include its energy in the EFE: $T^{\mu\nu}=T_1^{\mu\nu}+T_2^{\mu\nu}+\cdots$, where each $T_i^{\mu\nu}$ is a different source of energy. But we dont include a term $\boldsymbol{T_i^{\mu\nu}}$ for curvature: gravity has no energy tensor:
$$ \text{r.h.s}=T_\text{EM}+T_\text{matter}+T_\text{quantum?}+\cdots+ \overbrace{ T_\text{gravity}}^{\text{NO!}} $$
There is no tensor for gravitational energy and there is no term for gravity in the r.h.s. of the the EFE. In the r.h.s. of the EFE you only include nongravitational forms of energy.
Note that when in vacuum (i.e. no matter/no radiation), the EFE are $$ R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=0 $$ You can clearly see that there is no term in the r.h.s., even though there is gravity.
Therefore, the answer to your first question is: the EMT should not contain gravitational energy.
Second Question
In GR it doesn't make sense to speak of gravitational energy.
Short explanation: as you probably know, in GR gravity is not a true force$^1$. Therefore, there is no potential energy$^2$ associated to it.
Longer explanation: in GR energy is defined as $$ T^{\mu\nu}=\frac{-2}{\sqrt{|g|}}\frac{\delta \mathcal L_m}{\delta g^{\mu\nu}} $$ where $$ \mathcal L=\sqrt{|g|}\left(R+\mathcal L_m\right) $$ is the (full) Lagrangian of the theory, and $\mathcal L_m$ is the nongravitational part of the Lagrangian. As you can see, the energy is defined through the nongravitational part of the Lagrangian, which in turns means that it simply makes no sense to speak of the gravitational energy.
Note that it is possible to define a quantity that (supposedly) looks a lot like a gravitational energy$^3$: $$ t^{\mu\nu}=\frac{1}{2}Rg_{\mu\nu}-R_{\mu\nu}+\frac{1}{2g}[g(g^{\mu\nu}g^{\alpha\beta}-g^{\mu\alpha}g^{\nu\beta})]_{,\alpha\beta} $$ but it is not in general accepted as a true gravitational energy (it's more like a formal analogy, without much use AFAIK). Its not even a tensor. However, to see how it works, we can calculate $t^{00}$ in the Schwarzschild solution in the weak field (non-relativistic) limit: $$ g^{00}=-\left(1-\frac{2M}{r}\right) \qquad g^{ij}=\delta^{ij} $$ (where now I take $c=G=1$)
The "potential energy" of this metric is $$ t^{00}=\frac{1}{2g} \nabla^2 (g^2)=\nabla^2g+\frac{1}{2g}(\nabla g)^2 $$ where $g=g^{00}$. I wonder if this looks like a potential energy to you (IMHO, it does not).
EDIT: There is a certain point I'd like to discuss a bit further. It just happens that in EFE we don't include curvature-energy in the r.h.s. You point out that if this were the case, then the curvature would cause more curvature, and the new curvature would cause even more, etc. Is this possible? or is it just absurd?
Well, I think you will like a lot this lecture by Feynman (Vol II. Chap 23: Cavity Resonators). The r.h.s. of Maxwell equations include both $\boldsymbol E$ and $\boldsymbol B$, so in this case there exists the feedback you talk about: a certain electric field can be responsible for a magnetic field, which in turns generates more electric field, which is responsible for more magnetic field, ad infinitum. But the result is a convergent series, so everything works out just fine (I know this is kinda irrelevant for your question, but I think its neat, and I believe you might like it as well :) )
$^1$: "In general relativity, the effects of gravitation are ascribed to spacetime curvature instead of a force." - from Wikipedia.
$^2$: "In physics, potential energy is the energy that an object has due to its position in a force field [...]" - from Wikipedia.
$^3$: this quantity satisfies certain relations that energy usually satisfies, and it's constructed using just the metric tensor $g$.