Non-trivial scalar quantity Is there any scalar quantity made of only the Christoffel symbols, determinant of a metric and tensors, not derivatives? In other words, can we construct a scalar quantity which cannot be written in terms of a covariant vector and tensors?   
 A: Your question can be rephrased as a question on whether we can construct a scalar from the metric $g_{\mu \nu}$ and its first derivatives $g_{\mu \nu,\kappa}$ (being able to use the usual algebraic operations such as trace or determinant). The answer is that the only true scalar constructible this way is the trace of the metric $g^{\mu \nu} g_{\mu \nu} = 4$.
This can be shown e.g. in the Riemann normal coordinates around the investigated point. In those coordinates $x^\mu$, the invesigated point is at $x^\mu=0$ and the metric assumes the form
$$g_{\mu \nu} = \eta_{\mu \nu} - \frac{1}{3} R_{\mu \alpha \nu \beta} x^\alpha x^\beta$$
I.e., the Riemann normal coordinates can be constructed around any point so that $g_{\mu \nu, \kappa}=0$ and $g_{\mu \nu}=\eta_{\mu \nu}$. But $\eta_{\mu \nu}$ gives nothing interesting apart from the trace. Since the definition of a true scalar is that a construction in any coordinate system should yield the same result, we now see that the only non-zero scalar is the trace of the metric AKA a trivial constant. (The determinant of the metric is obviously not a true scalar.)
In other words, there is no interesting scalar constructed from the metric and its first derivatives, truly invariant information about the gravitational field is only introduced with the second derivatives of the metric, more specifically the information is exclusively hidden in the Riemann curvature tensor.

See the notes of Leo Brewin on Riemann normal coordinates for their construction.
A: Are you familiar with the Lagrangian formulation of GR? What we do is calculate the Christoffel symbols $\Gamma$ using a metric $g$ and then using a metric $\tilde{g}:=g+\delta g.$ It can be shown that the variation 
$$\delta\Gamma=\tilde{\Gamma}-\Gamma$$
is an element of the tensor algebra $\mathcal{T}^1_2$. It is then trivial to construct a scalar from this tensor. I'm sure I could come up with more scalars, given enough time.
EDIT: $$\frac{\delta^4(x-y)}{\sqrt{|g|}}$$is also a scalar, but it has a delta function, so I'm not sure that's what you're looking for.
