I asked the question. However, I think I have a simpler example than the currently accepted answer.
Let $X=(1,0)$ and $Y=(0,1)$ be the constant unit vector fields in the $x$ and $y$ directions. Define a connection such that:
$$\begin{matrix}\nabla_XX=Y&\nabla_XY=-X\\
\nabla_YX=0&\nabla_YY=0\end{matrix}$$
Equivalently, $\Gamma_{12}^1=-1$, $\Gamma_{11}^2=1$, and all other Christoffel symbols are zero. Clearly, it is not symmetric.
Parallel transport of a vector along a curve is equivalent to rotating the vector by an amount equal to the horizontal displacement of the curve. Since rotation is an isometry of tangent spaces, this property tells us that the connection is compatible with the metric.
Since $\nabla_YY=0$, vertical lines are geodesics. The other geodesics each look like a component of the graph of $\ln(\cos(x))$, but translated some amount horizontally and vertically. (This can be derived from the aforementioned property of parallel transport.) In other words, the nonvertical geodesics look like components of the graph of $\ln(\cos(x+C_0))+C_1$ for some $C_0,C_1$.
To check this, we can take $C_0=C_1=0$ without loss of generality. We need a unit-speed parametrization of $y=\ln(\cos(x))$. It can be checked that the following is such a parametrization:
\begin{align}x(t)&=\tan^{-1}(\sinh(t))\\
y(t)&=-\ln(\cosh(t))\end{align}
The equations for a geodesic, in this case, demand:
\begin{align}\frac{{\rm d}^2x}{{\rm d}t^2}-\frac{{\rm d}x}{{\rm d}t}\frac{{\rm d}y}{{\rm d}t}&=0\\
\frac{{\rm d}^2y}{{\rm d}t^2}+\left(\frac{{\rm d}x}{{\rm d}t}\right)\!^2&=0\end{align}
Calculation shows that $\frac{{\rm d}x}{{\rm d}t}=\operatorname{sech}(t)$ and $\frac{{\rm d}y}{{\rm d}t}=-\tanh(t)$, and the result follows.
EDIT: There's another interesting property of this connection. If we define $\nabla'$ such that ${\Gamma'}_{ij}^k=\frac12(\Gamma_{ij}+\Gamma_{ji})$, then we end up with a symmetric connection with the same geodesics. However, parallel transport changes, and it is no longer compatible with the metric. In fact, $\nabla'$ is not compatible with any metric!
To see this, note that there is no geodesic that goes between $(0,0)$ and $(\pi,0)$, or between any pair of points whose horizontal distance is more than or equal to $\pi$. If $\nabla'$, which is symmetric, were compatible with any metric, then it would be the Levi-Civita connection on the corresponding Riemannian manifold. This would contradict the Hopf–Rinow theorem, which states that, for any complete Riemannian manifold, there exists a geodesic between any two points.
(This also means that the Hopf–Rinow theorem crucially uses the symmetry of the connection, since it fails with our asymmetric $\nabla$ as well.)