To understand the existence and uniqueness of the LC connection, it is not possible to sidestep some algebra, namely the fact (with a 1-line proof) that a tensor $a_{ijk}$ symmetric in $i,j$ and skew in $j,k$ is necessarily zero. The geometrical interpretation is this: once one has the $O(n)$ subbundle $P$ of the frame bundle $F$ defined by the metric, there exists (at each point) a unique subspace transverse to the fibre that is tangent both to $P$ and to a coordinate-induced section $\{\partial/\partial x_1,\ldots,\partial/\partial x_n\}$ of $F$.
This isn't a full answer but hopefully the beginning of one. I'm assuming it's a matrix group but it might work for all Lie groups anyway.
Let $G$ be a matrix group with Lie algebra $\mathfrak{g}$. Suppose $\gamma(t)$ is a geodesic in $G$ and $x(t)$ is the geodesic in the exponential coordinates, ie. $\gamma(t) = \exp(x(t))$. Then $x(t)$ satisfies an ODE $x''(t) + Q_{x(t)}(x'(t),x'(t)) = 0$, where, for given $x$, the $Q_x$ is some quadratic map $\mathfrak{g} \to \mathfrak{g}$. If we can find $Q$ then we just need to polarise it to a bilinear map $\mathfrak{g} \times \mathfrak{g} \to \mathfrak{g}$ and this bilinear map will contain all the Christoffel symbols.
There is a power series formula for the differential of the exponential map. See G. M. Tuynman, The Derivation of the Exponential Map of Matrices http://www.jstor.org/pss/2974511 - it implies:
$\frac{d}{d t}(\exp(x(t)) = \exp(x(t)) \left( \frac{1 - \exp(-\mathrm{ad}\ x(t))}{\mathrm{ad}\ x(t)}\right)x'(t)$
where $\mathrm{ad}\ x$ is the map which takes $y$ to $[x,y]$ and the fraction $\left( \frac{1 - \exp(-\mathrm{ad}\ x(t))}{\mathrm{ad}\ x(t)}\right)$ means $1 - \frac{1}{2!} \mathrm{ad}\ x(t) + \frac{1}{3!} \mathrm{ad}\ x(t)^2 - \frac{1}{4!} \mathrm{ad}\ x(t)^3 + \cdots,$ ie. it is the power series for the fraction as if $x(t)$ was a real number.
Because the covariant derivative is bi-invariant, the geodesic $\gamma$ must be of the form $\gamma(t) = \gamma(0) \exp(A t)$ for some $A \in \mathfrak{g}$. Differentiating $\exp(x(t)) = \gamma(0) \exp(A t):$
$\exp(x(t)) \left( \frac{1 - \exp(-\mathrm{ad}\ x(t))}{\mathrm{ad}\ x(t)}\right)x'(t) = \gamma(0) \exp(A t) A = \exp(x(t)) A.$
So $\left( \frac{1 - \exp(-\mathrm{ad}\ x(t))}{\mathrm{ad}\ x(t)}\right)x'(t) = A.$ Differentiating this,
$\frac{d}{dt}\left( \frac{1 - \exp(-\mathrm{ad}\ x(t))}{\mathrm{ad}\ x(t)}\right)x'(t) + \left( \frac{1 - \exp(-\mathrm{ad}\ x(t))}{\mathrm{ad}\ x(t)}\right)x''(t) = 0.$ So
$x''(t) + \left( \frac{1 - \exp(-\mathrm{ad}\ x(t))}{\mathrm{ad}\ x(t)}\right)^{-1} \frac{d}{dt}\left( \frac{1 - \exp(-\mathrm{ad}\ x(t))}{\mathrm{ad}\ x(t)}\right)x'(t)=0$, ie.
$\begin{align}x''(t) &+ \left( 1 - \frac{1}{2!} \mathrm{ad}\ x(t) + \frac{1}{3!} \mathrm{ad}\ x(t)^2 - \frac{1}{4!} \mathrm{ad}\ x(t)^3 + \cdots \right)\\\\&\quad\cdot \left( - \frac{1}{2!} \mathrm{ad} x'(t) + \frac{1}{3!} (\mathrm{ad}\ x'(t) \mathrm{ad}\ x(t) + \mathrm{ad}\ x(t) \mathrm{ad}\ x'(t)) - \cdots \right) x'(t) = 0\\\\\end{align}$
I'm not sure what to do now. We could compute a few terms of it to get some terms of $Q_x$, but it seems like there should be a nice formula, like the one above for the differential of $\exp$, or at least there should be a recursive formula for the coefficients of the series...? It's even possible that the formula simplifies WAY down, note eg that all the terms ending with $\mathrm{ad}(x'(t)) x'(t)$ die instantly.
Best Answer
Let $\nabla$ be the connection induced by the Levi-Civita connection. $\nabla$ is left invariant. It thus defines a bilinear product $b$ on ${\cal G}$ the Lie algebra of $G$. Let $c(t)$ be a geodesic. We can write $\dot c(t)=dL_{c(t)}(x(t))$ where $x(t)\in {\cal G}$. It you write the equation $\nabla_{\dot c(t)}\dot c(t)=0$, you obtain
$\dot x(t)+b(x(t),x(t))=0$.