Interesting question. The answer is no: surfaces with bounded geometry can have normal
bundles with unbounded curvature.
To set the stage, it's worth first noting that you can have a surface with extreme geometry isometrically embedded in $\mathbb E^3$, where the normal bundle, being one-dimensional, has a trivial connection; or include this into $\mathbb E^4$ (Euclidean 4-space) where the
normal bundle is 2-dimensional, but the curvature is still 0. This at least illustrates that bounded geometry of the normal bundle and tangent bundle are decoupled.
I'll now describe an isometric embedding of the $\mathbb E^2$
into $\mathbb E^6$ where the connection
on the normal bundle has unbounded curvature. The embedding have local 1-parameter groups of symmetry, which makes it easier to keep track of curvature without needing to write down equations.
Start by visualizing
a helical curve in $\mathbb E^3$. The tangent vector to a helix goes repeatedly around a circle in its spherical image. The connection on the tangent bundle is induced by this
Gauss map from the connection on $S^2$, so the parallel translation of the normal bundle rotates the plane by an angle equal to the area enclosed inside this circle once every coil of the helix.
Now consider a similar curve in $\mathbb E^5$, thought of as $\mathbb E \times \mathbb E^2 \times \mathbb E^2$. In the $\mathbb E$-direction, the curve makes uniform progress, while going around circles of possibly different radii at possibly different rates in the $\mathbb E^2$ directions. If the term weren't otherwise engaged, one could call this a double helix.
It is invariant by a 1-parameter group of isometries of $\mathbb E^5$ that translates in
the $\mathbb E$ direction while spinning the two perpendicular planes at their own rates.
The normal bundle splits into two $\mathbb E^2$ subbundles, its intersection with
the two 3-dimensional $\mathbb E \times \mathbb E^2$'s. The connection preserves this splitting, rotating the two $\mathbb E^2$'s indendently.
Now add an extra "parameter" dimension, making the ambient space $\mathbb E^6$.
Modify the curve in $\mathbb E^5$ by increasing the radius of one helix while decreasing the
radius of the other, balancing the changes so the curve remains invariant by the same
1-parameter group, and its arc length remains constant (as measured by the time parameter of the 1-parameter group). It's easy to see, since it's
locally isometric to a surface of revolution because of the symmetry, that the
resulting surface is isometric to $\mathbb E^2$. We can make the circle in on
$\mathbb E^2$ go all the way to 0. After making sure it has $C^\infty$ contact to
a straight line in this projection, we can then start making this projection helical
again, but with a
steeper and tighter helix, adjusting by letting the other helical projection shrink to a line. The local symmetry group in $\mathbb E^6$ has changed, but the induced symmetry on
the surface remains the same.
We can go back and forth, alternating between helical effects in the two factors,
inexorably tightening the screws without distoring the surface.
The curvature of the induced connection on the normal bundle becomes arbitrarily high,
as you can see by following the connection around a small rectangle, with two edges
in the "parameter" direction, one edge where say the first helix has become straight
and the fourth edge where the first helix is wound in very small tight coils.
Best Answer
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.