Question: I am asking this in the context of differential geometry (specifically Riemannian).

When the Levi-Civita Connection is defined, we require that the torsion tensor is 0, which in local coordinates translates to the requirement that $\Gamma\_{ij}^{k} = \Gamma\_{ji}^{k}$; which is the covariant derivative version of saying partial derivatives commute: $\nabla\_{\partial\_i}(\partial\_j)=\nabla\_{\partial\_j}(\partial\_i)$.

This is obviously true in the Euclidian settings, and I understand all the details of the proofs. But why is this such an essential property? Why does this capture our intuitive sense of derivatives?

Answer: 
The covariant derivative version of trying to commute partial derivatives is: $\nabla\_{\partial\_i}\nabla\_{\partial\_j}-\nabla\_{\partial\_j}\nabla\_{\partial\_i} - 0 =
R(\partial\_i,\partial\_j)$.

Torsion is measuring something different: It is the covariant derivative of the soldering form
$\sigma\in\Omega^1(M,E)$ which you use to identify the vector bundle $E$ with $TM$, where $E$ is the bundle you are considering your covariant derivative on.
The Levi-Civita connection is just a very special one - torsion free. It is interesting that the same geometry may be described by switching to another, non-torsion-free connection. E.g. there is such a version of general relativity which is called teleparallel formulation. While the curvature tensor (based on the new connection) vanishes, all the deviation from flatness has been shifted to the torsion tensor (better: vector-valued 2-form). Einstein exchanged his ideas with Cartan in the 1920 about that.
Torison has also an equivalent in physics as dislocation density (disclocations are defects in crystals). The theory has been developed in the 1950's by Kondo, Bilby and Kröner. See also the book Ricci calculus by J.A. Schouten.
To summarize, it is not that important that the connection is symmetric, it is merely a matter of choice. The metric, for instance, is independent of the connection.