While we don’t know the answer to this there are theories involving inflation or vacuum energy that could provide an explanation.
https://www.mso.anu.edu.au/~charley/papers/Chapter22Lineweaver.pdf
Perhaps a more intriguing question where there is a definitive answer is why the entropy of the universe increases with time in the first place.
As highlighted in the link, the universe is expanding in a thermodynamic
isentropic process where entropy is conserved and not increase with time.
The solution to this dilemma is to examine how temperature scales during different epochs in the Universe’s history.
The scale factor “a” is defined a = R(t)/R(t₀) where R(t) is the cosmic scale in the past at some time cosmological t.
R(t₀) = 1 is the cosmic scale now.
The density of matter ρ = m/V (mass/volume).
V ≡ a³ hence ρ scales as a⁻³.
Energy density for matter εₘ = mc²/V also scales as a⁻³.
In the very early universe however which is radiation dominated the energy density εₓ is;
εₓ = hν/V = hc/Vλ where λ is the wavelength of the photon.
Since V ≡ a³ and λ ≡ a then Vλ ≡ a⁴.
The energy density of radiation εₓ scales as a⁻⁴.
Temperature scales differently in a radiation dominated and matter dominated universe.
The
Stefan-Bolzmann law tells us the power radiated j by a blackbody is proportional to the 4th power of its temperature T.
j ∝ T⁴.
j = E/t = εₓV/ t ≡ a⁻⁴a³ ≡ a⁻¹
Hence in a radiation dominated universe temperature
Tₓ ∝ a⁻¹ and Tₓ scales as a⁻¹.
In a matter dominated universe we can model the universe as an ideal gas undergoing expansion against a constant external pressure p.
The gas is also travelling at non relativistic speeds.
Since the universe is undergoing adiabatic expansion the energy change dE is;
dE = -pdV where dV is the change in volume.
In terms of the scale factor;
d(a³ εₘ) = -pd(a³)
Since the gas is ideal;
p = nkTₘ where n is the number of particles and k is the Boltzmann constant.
Statistical mechanics tells us the average energy density εₘ = (3/2)kTₘ, hence the average pressure is;
p = (2/3)εₘ
Combining this with the equation p = nkTₘ gives;
εₘ = (3/2)nkTₘ
The total energy density needs to include the mass terms of n particles (= nmc²) hence;
εₘ = nmc² + (3/2)nkTₘ
Substituting εₘ = nmc² + (3/2)nkTₘ and p = nkTₘ into the equation d(a³ εₘ) = -pd(a³) gives;
nmc²d(a³n) + (3/2)kd(na³Tₘ) = -nkTₘd(a³)
nmc²d(na³) + (3/2)ka³[d(nTₘ) – Tₘd(na³)] = -nkTₘd(a³)
Note na³ is the number of particles in a volume V.
Since the number of particles in the volume is constant d(na³) = 0 the equation can be simplified to;
(3/2)ka³d(nTₘ) = -nkTₘd(a³) or;
(3/2)dTₘ/Tₘ = -d(a³)/a³
Solving the equation;
log(Tₘ) = -(2/3)logₑ(a³) + logₑ(A) = logₑ(a⁻²) + logₑ(A) or;
Tₘ = Aa⁻²
This differential equation is of the form
Tₘ ∝ a⁻² hence in a matter dominated universe temperature scales as a⁻².
Since the current universe is composed of both radiation and matter, thermal equilibrium and maximum entropy can only be achieved if Tₓ = Tₘ.
However since Tₓ and Tₘ scale differently and Tₘ cools down more rapidly in an expanding universe maximum entropy not been achieved since Tₓ ≠ Tₘ, but continues to increase despite the isentropic process.