Just based on the quadratic drag of air, yes, the fired bullet would take longer to hit the ground.
Just consider the vertical force caused by the air friction:
$F_y = - F_{\rm drag} \sin \theta = - C (v_x^2 + v_y^2) \frac{v_y}{\sqrt{v_x^2 + v_y^2}} = - C v_y \sqrt{v_x^2 + v_y^2}$
Where $\theta$ is the angle above the horizon for the bullet's velocity, and $C$ is some kind of drag coefficient. Note that when the bullet is moving down $\theta$ is negative, as is $v_y$, so the overall vertical force is positive and keeps the bullet off the ground for slightly longer.
In the dropped case, $v_x = 0$, so we get $F_y = -C v_y^2$.
In the fired case, we can neglect $v_y$ in the radical (assuming it's much smaller than $v_x$) and we get $F_y \approx -C v_y |v_x|$.
In other words, the upward force on the fired bullet is stronger, by a factor of $v_x / v_y$.
So freshman-level physics is wrong, at least according to sophomore-level physics.
Bonus Case:
If you're assuming a flat surface on earth, it's worth considering that many "flat" things (like the ocean) actually curve down and drop off below the horizon. In case you want to account for this curvature, it may be worth going to the bullet's reference frame with $\hat{y}$ always defined to point away from the center of the earth. Note that this puts you in a rotating reference frame, and then look at the centrifugal "force":
$F_y = m a = m R \omega^2 = m R \left(\frac{v_x}{R}\right)^2 = m \frac{v_x^2}{R} $
Where $R$ is the radius of the earth and $m$ is the mass of the bullet. So again, an upward force, this time proportional to $v_x$ squared. Of course this is the same as pointing out that the earth curves away from a straight line, but it's another fun application of not-quite-freshman physics.
Now you can add in much more complicated aerodynamics, but there the question sort of looses its undergrad physics charm there and becomes an aerospace engineering question!
Other answers don't mention the fact that no single impulse (e.g, like being fired from a gun) can launch a projectile into orbit. A purely ballistic projectile fired from a gun must either crash back into the planet, or it must escape from the planet altogether.
In order to achieve orbit, at least two impulses must be applied to the projectile. The first one (from the gun) launches it into an elliptical trajetory that returns to the surface, and then the second impulse must be applied by a rocket motor to "circularize" the orbit at the moment when the projectile reaches the apogee of the initial ellipse.
Best Answer
As far as I know, there is no data to calculate the drag effects of the atmosphere at these speeds.
The Pascal-B shot of Operation Plumbob did, apparently, launch a 1-ton steel plate at 6 times escape velocity. https://en.wikipedia.org/wiki/Operation_Plumbbob
Nobody has the faintest idea of whether or not it actually made it out of the atmosphere, although the most likely result is that it vaporized.