Well, jet engines do have gyroscopic effects. It is a major concern in the design of the turbomachinery. When the plane pitches/yaws, the resulting gyroscopic moment causes the compressor and turbine blades to move closer to the case. If excessive, this can cause the blades to rub into the case, causing loss of performance.
As to why the plane is not affected by the gyroscopic moment of the engines, let's compare a few numbers.
Take a typical WWI era plane such as the Vickers FB19 powered by a Le Rhone 9J. The engine has a mass of 146 kg, the majority of which is rotating and produces 110 HP. The plane has a maximum takeoff mass of 675 kg, which includes the engine. The rotating mass is a fairly sizable fraction (~20%) of the mass of the plane. Therefore the gyroscopic effect of the engine has a big impact on manueverability.
Now compare that with an F16 powered by an GE F110 engine. The high pressure spool of the engine has a mass on the order of 200 kg (the whole engine is about 2000 kg and the high speed rotating spool is about 10% of that... sorry I don't have a specific reference). The max takeoff mass of the plane is about 19,200 kg. The rotating mass is only 1% of the total mass of the plane. So the gyroscopic effect is not so important. i.e. compared to the moment required just to turn the plane itself, the gyroscopic moment is not large.
Further, to be more precise, the gyroscopic moment is not actually proportional to the mass, but to the mass moment of inertia, which is proportional to radius squared. A lot of the rotating mass of an older rotary engine was at a fairly high radius, whereas a lot of the rotating mass of a jet engine rotor is at a fairly low radius. Although I have not done the calculations, my guess is that a modern fighter jet engine spool probably has a lower moment of inertia than a WWI rotary engine, despite the higher mass.
In response to the comment by J Walters, let me try to make this a little more precise. The moment required to perform an angular acceleration is $M=I\alpha$, where $I$ is the moment of inertia. Let's assume that the plane is a rod (length >> width or height). That's not totally correct, but it's good for an order of magnitude approximation. Then from this formula, $I=(1/12)mL^2$. So for the FB19, the moment of inertia is $(1/12)(675)(5.54^2)=1726 kg-m^2$. For the F16, $(1/12)(19200)(15.06^2)=362885 kg-m^2$. So, comparing these two, the moment that the control surface have to apply to turn the plane (at a given angular acceleration) is 200x higher for the F16.
Now, let's look at the moments due to the gyroscopic effect. The gyroscopic moment is $M=J \Phi \times \Omega$, where $\Phi$ is the angular velocity of the aircraft maneuver and $\Omega$ is the spin angular velocity of the engine, and $J$ is the polar moment of inertia. I said before that I expected $J$ for the F16 to actually be lower due to lower radius, but let's just assume they are the same. The Le Rhone spins at 1,350 RPM. I don't know exactly, but I know the F110 high pressure spool top speed is between 15,000 - 20,000 rpm. So the gyroscopic moment is about 10 - 15x higher for the F16.
And as Peter Kampf pointed out, the aerodynamic forces go up too. The FB19 had a top speed of about 100 mph whereas the F16 can hit 900 mph at sea level. From NASA's excellent series, both lift and drag scale with velocity squared. So the aerodynamic forces are 81x higher for the F16.
So, in summary, yes, the gyroscopic moment is one order of magnitude larger on the F16. But everything else that the plane has to deal with during maneuvers (the moment of inertia of the plane itself, aerodynamic forces), are two orders of magnitude higher. So the gyroscopic forces are less relevant in comparison.