Nope, there is no difference in free space attenuation.

From a previous thread...

True, there is no energy "lost in space"; its dispersed over a larger
area such that the flux falling on a given area is smaller. It is
frequency dependent only in the sense of antenna aperture; not space loss.

That would be the near earth terrestrial loses or attenuation. During
a seasonal test, we noted greater than expected losses at 5.8 Gig and
the predicted losses at 2.4 Gig. Several weeks later when the prairie
grass approached 2.5 inches, the loss at 5.8 Gig was as predicted,
but the loss at 2.4 was greater than before and greater than predicted.
Generally, we see greater losses at 5.8 when shooting through trees,
but then we make sure our antennas are well over mid-point tree lines.

True, they do not relate as I said "That applies with true free space
(earth to satellite) or greater than several Fresnel zone clearances,"
where an earth to satellite would not even have an Fresnel zone excursions.
But with typical cellular communications you do indeed have Fresnel zone
incursions.

Posted sometime ago in another NG:

In the past I argued the higher the frequency, the greater the
attenuation, but failed to note I was referring to terrestrial
propagation...but I'll get to that later.

Lets look at the traditional free space loss formula:
Path loss in dB = 32.4 + 20 log(f) + 20 log(d), where f is frequency
in MHz and d is distance in miles.

It implies increasing the frequency will increase the path loss
(greater attenuation). When you run a plot at 2.4 GHz and one at 5.8
GHz, you'll find there is 7.7 dB more loss at 5.8 GHz.

[added for clarity...
However its the decreased antenna aperture that accounts for this
"lose", not free-space attenuation.]

If you go backwards in the equation and see how it is derived, you'll
find the capture area (antenna aperture) defined as wavelength squared
divided by four times Pi.

For example - Lets take two simple dipoles for 2.4 GHz and 5.8 GHz
with the ends of the dipoles at 2.45 inches and 1 inch apart
respectively. Using the above formula, we find the capture areas are
1.912 and .318 square inches. Divide the area of the first antenna by
the second antenna's area and you'll get 6, or six times smaller (so
you would express it as a negative 6)

Convert -6 to dB and you get -7.7, the same loss number you go in the
plots you ran above.

Therefore, there really isn't greater attenuation as you increase the
frequency, rather the antenna is "less sensitive".

---------------------------------------

Now...on to terrestrial losses.

I placed a 2.4 GHz and a 5.8 GHz transmitter with simple vertical
dipole antennas on the output connectors at 500 feet up a tower and
measured the signal level from another 200 foot tower nearby with
simple vertical dipole antennas. This is as close as you'll get to
true free space on the face of the earth. As expected, the 5.8 GHz
signal was 7.7 dB lower. I had similar signal differences from several
test points with antennas on a forty foot mast and a clear line of
site.

I ran the tests again at several locations in the county with the
transmitters on a forty foot mast and ten foot mast for the receiver
antennas. I saw some additional loss from trees as expected and a
greater loss at 5.8 GHz. The additional loss varied with the terrain
and foliage. The conclusion would be there are greater terrestrial
losses at higher frequencies.

Eventually I'll get a system set up where I can measure hourly signal
levels from several other WISP's access points from miles away over a
year period. I suspect I'll see the same results my casual testing has
shown...path fading is worst for a few weeks in the spring and a few
weeks in the fall, and about an hour after sun up after the sun warms
the earth and you have a layer of warm air under the cooler morning
air, and an hour before sun down when the air starts to cool off and
drift into low spots under the still warm evening air.