Can Lagrangian have a potential term proportional to the quadratic or higher of velocity?
The EL equations for your proposed potential would be
$$m\ddot q -\frac{d}{dt}\frac{\partial U}{\partial \dot q}= \frac{\partial U}{\partial q}$$ which can be re-expressed as $$\left(m- \frac{\partial^2 U}{\partial \dot q^2} \right)\ddot q = -\frac{\partial U}{\partial q} +\frac{\partial^2 U}{\partial q \partial \dot q} \dot q$$
Non-linear dependance of $U$ on $\dot q$ would therefore result in a effective dynamical mass term, while cross-dependance on $q$ and $\dot q$ would yield a velocity-dependant force, which could describe some form of fluid friction or the Lorentz force, as you mention in the original question.
It should be noted, however, that the addition of such velocity dependance could make the Hessian matrix $\frac{\partial ^2 L}{\partial \dot q^i \partial \dot q^j}$ singular. In this case, the transition to the Hamiltonian framework via Legendre transfom must be approached with much more caution - see e.g. this question.
Consider the relativistic Lagrangian
$$L = mc^2 \sqrt{1-\frac{v^2}{c^2}} - U(\mathbf x)$$
Expanding to second order in $v^2/c^2$, this becomes
$$L = \frac{1}{2} mv^2 +\frac{3}{8} m v^4/c^2 - U(\mathbf x)$$ which yields the EL equations $$-\frac{\partial U}{\partial \mathbf x} = \frac{d}{dt}\left[m\left(1+ \frac{3v^2}{2c^2} \right)\mathbf v\right]$$ This is Newtonian mechanics, except that $m$ has been replaced by $m\left(1+\frac{3v^2}{2c^2}\right)$, giving the lowest-order relativistic correction.
For what it's worth, higher powers of velocities (or momenta) do appear in many Lagrangians, e.g. as a power series in relativistic corrections, say, from square root terms.
In fact, there are a lot of systems, whose Lagrange contains higher (even infinite) order of velocity in non-relativistic mechanics, for example, if you have a simple equation of motion $$ \ddot q+c_1 \dot q=0. $$ Then the Lagrange can be written as $$ L= \dot q f(\dot q) \exp(-c_2 q) $$ where $f(\dot q)$ satisfies the equation $$ \frac{d f}{d\dot q}=\frac{1}{\dot q^2}\exp\left(-\frac{c_2}{c_1} \dot q \right) $$ where $c_1$ and $c_2$ are two constants.