Is the gravitational potential a measurable physical quantity or an artifact of warped measures? The Euler-Lagrange conditions for stationary points of $$L=m/2 v(\mathbf{\dot{x}})^2-U(\mathbf{x})$$
($m$ is mass, $v()$ is velocity, $U()$ is the scalar potential, and the boldfaced arguments of these functions are vectors)
are given by
$$-\nabla U = \mathbf{\ddot{x}}m$$
My amateurish understanding is that if the scalar potential is gravity, then in all local frames $U(\mathbf{x})$ is in fact a constant, i.e. independent of $\mathbf{x}$ (the particle is said to be "free"), in which case the E-L conditions in the gravitational force field (assuming no other forces are present) reduce to:
$$\mathbf{0}= \mathbf{\ddot{x}}m \tag{1}$$
But that gravity nonetheless makes itself felt by acting on the metric tensor, so this solution is not useful because it does not take account of changes in the metric tensor. The useful solution, taking account of the metric, is (switching to tensor notation),
$$-\Gamma^a_{ij}\dot{x}^i\dot{x}^j=\ddot{x}^a \tag{2}$$
However I see here that it is perfectly acceptable to treat the gravitational scalar potential as a physical quantity dependent upon position, and hence ok to write things like
$$\mathbf{g}= \mathbf{\ddot{x}}m \tag{3}$$
(where in the notation of this post $\mathbf{g}=-\nabla U < \mathbf{0}$)
instead of equation 1 or 2.
So which is it? Is the gravitational potential a physical quantity analogous in its measurement to other scalar potentials like temperature; or is it an artifact of the warping of the measurement frame itself? Or is there no meaningful distinction between these two characterizations?
When we see $\mathbf{g}$ are we supposed to understand
$$-\Gamma^a_{ij}\dot{x}^i\dot{x}^j=g^a$$
?
 A: The Lagrangian that you wrote is just an approximation. That is, in fact, what the gravitational potential really is - an approximation, albeit a really good one.
The Lagrangian of a free particle in General Relativity is the following:
$$ \mathcal{L}(x^{\mu}, \dot{x}^{\mu}) = - m c \sqrt{g_{\mu \nu}(x) \dot{x}^{\mu} \dot{x}^{\nu}}. $$
Here $\{x^{\mu}(\tau)\}$ is a parametric form of the particle's worldline, and $\tau$ is an arbitrary parametrization of that worldline, and $\dot{f}$ is the derivative of $f$ with respect to $\tau$ (and not the coordinate time $t = x^0(\tau)$!).
Changes in parametrization of the form $\tau \rightarrow \tau'(\tau)$ for an arbitrary function $\tau'$ result in the Lagrangian changed by a total derivative (a good exercise is to demonstrate it). Thus, reparameterizations are a gauge symmetry of this Lagrangian.
The resulting equations of motion are quite complicated, however, by making use of the gauge symmetry, they can be written in a simple form which corresponds to choosing $\tau$ to be the proper time along the worldline. In this gauge, the equations of motion are
$$ \ddot{x}^{\mu} = - \Gamma^{\mu}_{\nu \sigma} (x) \dot{x}^{\nu} \dot{x}^{\sigma}, $$
which is precisely the geodesic equation. It actually makes a lot of sense, because the action functional (integral over $\tau$ of the Lagrangian) measures precisely $m c$ times the proper time passed according to the particle's internal clock. Extremizing the proper time leads to the analog of the straight line in curved spacetime - the geodesic.
A: 
When we see $\mathbf{g}$ are we supposed to understand $-\Gamma^a_{ij}\dot{x}^i\dot{x}^j=g^a$?

No, because the quantity $-\Gamma^a_{ij}\dot{x}^i\dot{x}^j$ is not just a function of position, it also depends on the velocity $\dot{x}$. Also, because $\dot{x}$ means a derivative with respect to an affine parameter $\lambda$, the quantity you're talking about implicitly depends on the choice of affine parameter as well. The choice of an affine parameter is arbitrary up to a linear transformation. Although it's true that for a timelike geodesic it's natural to take the affine parameter to equal the proper time, there is no such natural default for null geodesics.
The most closely analogous thing to the Newtonian $\textbf{g}$ is simply $-\Gamma$.

Is the gravitational potential a measurable physical quantity or an artifact of warped measures?

Neither. Most spacetimes cannot be defined in terms of a gravitational potential $U$. The closest counterpart to a gravitational potential in general relativity is the metric, but the metric is a tensor, not a scalar like $U$.
Although your question asks for the GR analog of $U$ and whether it's measurable, it's also worth pointing out that the GR analog of $\textbf{g}$ is not measurable. The Christoffel symbol $\Gamma$ can always be made to vanish at a point by a proper choice of coordinates. This is basically an expression of the equivalence principle.
What's measurable in GR is the Riemann tensor (or other measures of curvature constructed from it and its covariant derivative). The Newtonian analog of the Riemann tensor would be some measure of tidal stress.
