Aren't the interference fringes which make up the hologram much
smaller than wavelength of light?
Nope, they're a bit larger.
Example: two beams crossing at 90deg, insert a film plane, and the fringe spacing will be sqrt(2) of the wavelength, or 1.41x longer. (That's with the film plane perpendicular to the line between the two beams.) If the angle between the beams is smaller, the fringe-spacing will be larger. The easiest holograms to record will position the object very close to the reference-beam origin, so the angles remain very small, and the fringes very wide. That's how those "holospex" glasses work, the ones converting your night-time xmas-lights bokeh into snowflakes or dollar-signs. Those are small-angle conventional holograms, so the color smearing remains fairly small.
Even if we had LCDs with massive resolution, wouldn't the diffraction limit prevent the using of them to pattern the plate
Not with contact-printing, or when using UV to duplicate a visible-light diffraction pattern. Or, see below, step-repeat.
I've never seen a straightforward explanation of how computer
holograms are actually transferred to the physical recording medium.
In old CIRCUIT CELLAR magazine they had a hobbyist CGI hologram project.
You printed out your synthetic hologram as a grid of many 300DPI laserprinter pages. Then photographed it onto slide film, with DPI being more than 100x higher than 300. In your "virtual holo" equations, as long as the angular size of the hologram and the ref-beam angles were small (and NOT 90deg as above,) then the resulting fringes on the developed film were at least a few wavelengths across. I never tried that project, but did notice that an 8-1/2in page would end up as about 2mm wide on the slide transparency.
All the above explanation applies to conventional holograms, the "off-axis holography" type, which requires lasers for viewing. Benton Rainbow holograms are significantly different, since Benton's trick makes them frequency-independent, and therefore size-independent. We actually can scribe the individual fringes of a White Light hologram onto a flat sheet, then view the resulting 3D image in sunlight or a bright pointsource illuminator.
I released this online, as a science-fair project. A decade later, several artists including Tristan Duke of MIT took it forward as an art form, and now they appear as rotating "abrasion holograms" on vinyl record albums, by Jack White, Rush 2112 re-release, and a Star Wars soundtrack release.
If this is the case, why does one not need an electron microscope
to etch the fringes?
The easy way is to simply use two beams to put interference patterns directly on film! This is "step-and-repeat" printed holography, where each pixel written onto the film is a diffraction grating created by two off-axis laser beams. Two simple beams would create a parallel grating. If one of the beams is actually the light reflected off an object (or off a low-res LCD,) then each pixel contains a diffraction grating which stores a 2D image of the entire object, as viewed from one particular position. To make a full hologram of a 3D object, let the LCD image change for each holo-pixel in the raster, so the LCD displays a 2D view of a 3D object as seen from different viewpoints (seen from each holo-pixel's position.) That's what conventional holograms are: like a "window screen" where each little square hole contains a large 2D view of a 3D object. Each "holo-pixel" or "window-screen hole" is a hologram, but it only stores a 2D view of the final object. Pull your face away from the "window screen," and the array of tiny square holograms is now a conventional hologram of a 3D object. (It's a bit like "lenticular" 3D devices. But in this case each tiny lens is instead a small hologram, laid down by step-and-repeat on an optical bench.
