I think you are essentially correct that the device does not run smoothly, but the importance of this effect will depend on things like the thickness of the wheels and how well-built the table is.
I will try to answer the following question: "What is the minimum force we must pull with on the outside rim to make the table spin?"
For this calculation we'll pretend that the construction is essentially perfect - the wheels are super-smooth, perfectly aligned, etc.
If the turntable is spinning, any given wheel is moving in a circle. An ideal wheel contacts the ground beneath it with just the very bottom of the wheel. This bottom portion is a line segment whose width is the width of the wheel. Let's call that the "contact line" and make its width $d$ (I want to use $w$ later for the weight).
Suppose we painted the rim of one of the wheels. Then as it went around in the circle, it would paint the ground beneath it with a ring whose thickness is $d$. Let's call the radius of that ring $R$. This is the same as the distance from the center of the device to the wheel.
Resistance to turning arises when the bottom of the wheel slips, or moves relative to the ground. For a wheel going in a straight line, this slipping could ideally be zero - the bottom of the wheel would not be moving at all relative to the ground (while the top of the wheel would be moving at double speed to make up for it.) When the wheel moves in a circle, though, there must be some slipping because different parts of the wheel's contact line move different distances.
When we go around one full circle, we want to know the total amount of slipping that occurs. We'll say that the very center of the contact line doesn't slip at all. The outside and inside edges move in larger and smaller circles respectively, so their total slipping distance is $d\pi$ each (in opposite directions, but that doesn't matter much). The average slipping distance across the wheel is $d \pi/2$.
Next we want to know what force friction with the ground exerts on slipping wheels. This is proportional to the weight of the turntable, $w$, and to an unknown coefficient of friction with the ground $\mu$. $\mu$ might be around one half or so.
The total work done is the force $w\mu$ times the slipping distance $2 d \pi$, so the work needed to rotate the ring one full rotation is
$$W = 2w\mu d \pi$$
(The number of wheels doesn't matter, because if we add more wheels, the weight on each wheel goes down proportionately.)
This work is also equal to the force we exert on the ring times the distance around, which in this case is $2 \pi R$. This means we can set
$$2 \pi R F = 2 w \mu d \pi$$
and simplify to
$$F = w \mu \frac{d}{R}$$
The wider the wheels, the heavier the table, the higher the friction, and the smaller the table, the harder it will be to turn the table.
Now we want to know whether this matters. Is this effect comparable in size to the rolling resistance we expect with wheels anyway?
The Wikipedia article linked above says that a typical car tire has a rolling resistance of $0.01$. Perhaps a well-constructed turn table could achieve the same. The comparable statistic from the sliding of the wheels is $\mu d/R$. A reasonable ratio of the wheel thickness to the radius is $1/10$, so if $\mu \approx 1/2$ we get that the resistance due to slipping is around $0.05$, perhaps five times as big as the rolling resistance.
The punch line is this: if the cart is exquisitely well-constructed and has very thick wheels, then the fact that the wheels don't align perfectly will be the dominant cause of resistance. If the cart is poorly-constructed, with sticky wheels, or if the wheels are very thin, then the slightly-misaligned wheels don't matter much. In my best guess for an average scenario, the extra resistance added by slipping wheels is, to an order of magnitude, about five times the ordinary rolling resistance, and the spinning turntable will be significantly harder to spin than a simple cart constructed the same way would be to push in a straight line.
