The altitudes are not what I would call ridiculous. Though you seem prone to using unjustified superlatives in your questions.
The optimum, i.e. minimum injection velocity sub-orbit was provided by HopDavid in this answer. With some manipulation, you get that the maximum altitude of the optimal trajectory is:
$${r\over 2}\left(\sin{\alpha\over 2}+\cos{\alpha\over 2}-1\right)$$
where $r$ is the radius of the injection and $\alpha$ is the central angle of the traverse. The optimal sub-orbits look like this:
The green lines show the central angle traversed, and the solid orange curve is the trajectory. The dotted orange curve completes the rest of the sub-orbit.
The maximum altitude is at a traverse of $\pi/2$, or a fourth of the way around the planet, about the distance between Los Angeles and Moscow (since we're talking about ICBMs). That maximum altitude is:
$${r\over 2}\left(\sqrt{2}-1\right)$$
or about $0.207\,r$. I would consider a reasonable injection altitude to be 50 km, so this gives a maximum ICBM altitude of 1330 km. That would nick the bottom of the inner van Allen belt, which starts at around $0.2\,r$. For most traverse angles you will be lower and should miss the belt. (Though watch out for the South Atlantic Anomaly if you're going from, say, Los Angeles to Port Elizabeth, South Africa.)
If you accept a penalty in the injection velocity, you can lower the peak altitude. For example, to reduce the peak altitude for a $\pi/2$ traverse by a factor of two to 665 km, you would need to increase the injection velocity from 7.17 km/s to 7.29 km/s.
