Sand is too viscous to flow like that...
The resistance of a fluid to forming waves is measured by the fluid's viscosity. Viscosity measures to ability of motion in one part of a fluid to impart motion to another part of the fluid. In the case of shear viscosity, it is the ratio of the shear stress to rate of shear deformation, or velocity of deformation. The viscosity of water is around 1 mPa-s at 20 Celcius.
Sand is not a fluid, so it doesn't properly have a viscosity. However, in situations like earthquakes, it can effectively flow like a fluid; indeed it can flow like a fluid in an hourglass too. So there are measurements of its effective viscosity.
Water molecules can each move around each other at the molecular level, sand grains are much much bigger. So the effective viscosity of sand is much more variable, since sand grains are different sizes and compositions, while water molecules do not differ (much). This book on earthquake engineering suggests viscosity values of around 100 to 1000 kPa-s for sand; this is 100 million to a billion times higher than water.
So the force needed to perturb a bucket of sand is a around a billion times higher than the force needed to perturb a bucket of water; similarly, the force needed to make a 10 foot tide in an ocean of sand is around a billion times higher than the force needed to make a 10 foot tide in the Ocean.
...anywhere on a planet
Can we imagine sand forming tides? Then we have to imagine a world with at least 6 orders of magnitude more tide-causing gravitational variation.
We can estimate the magnitude of tidal forcing from the force of gravity the tide-inducing object exerts on the planet in question. For example, the moon's gravitational effect on an object on Earth is $$\frac{F}{m_{\text{obj}}} = \frac{GM_{\text{moon}}}{r^2}$$ where Gm, the standard gravitational parameter, for the moon is $4.905\times10^{13} \text{ m}^3\text{s}^{-2}$, and r is the distance of the moon at 384399 km. We then get $F/m = 0.0003319 \text{ N/kg}$.
What if the tide causing object was much closer? The force of Jupiter's gravity on the moon Io is $$\frac{GM_{\text{Jupiter}}}{r^2} = \frac{1.267\times10^{17}}{421700000^2} = 0.712 \text{ N/kg},$$ which is about 3 orders of magnitude higher than seen on Earth. Much more powerful, but not quite there.
Lets see if we can get even better. The Roche limit (the limit to how close a satellite can be to the object it orbits) for the Earth and the Sun is 556,000 km, and the gravitational parameter for the sun is a hefty $1.327\times10^{20}$. If we set the Earth's distance from the sun to the Roche limit (Note: this is within the outer surface of the sun, so this is just theoretical demonstration), then we get $$\frac{1.327\times10^{20}}{556000000^2} = 429 \text{ N/kg}.$$
These tidal forces are now a million times stronger than those on Earth. Unfortunately, we have reached a limit. If the Earth got any closer to the sun, those powerful tidal forces would rip the planet apart. And they are still not powerful enough to make sand dune flow like water.
Conclusion
Tidal forces strong enough to make sand dunes flow in tides, would also be powerful enough to rip the planet they were on apart. Therefore, tides in sand oceans are impossible.