The confusion of entropy change
The standard free energy of formation of graphite is $0\ \mathrm{kJ/mol}$, and the standard free energy of formation of diamond is $2.9\ \mathrm{kJ/mol}$.  So, at $25\ \mathrm{^\circ C}$ and $1\ \mathrm{bar}$, graphite is favored thermodynamically over diamond.  Let's see what pressure would be required for $\Delta G$ to be zero at $25\ \mathrm{^\circ C}$.  Since the temperature is constant, the change in $G$ is described by $\mathrm dG=v\,\mathrm dP$, where $v$ is the molar volume.  The density of graphite is $2.26\ \mathrm{g/cm^3}$ and the density of diamond is $3.51\ \mathrm{g/cm^3}$.  So the molar volumes are $5.31\times 10^{-6}\ \mathrm{m^3/mol}$ for graphite and $3.42\times 10^{-6}\ \mathrm{m^3/mol}$ for diamond.  So, at pressure $P$ (in Pa), the Gibbs free energy for graphite and diamond are:
$$G_\text{Graphite}=0+5.31\times 10^{-6}P$$
$$G_\text{Diamond}=2900+3.42\times 10^{-6}P$$
So, $\Delta G$ at pressure $P$ is $$\Delta G_\mathrm{G\rightarrow D}=2900-1.89\times 10^{-6}P$$So the change in free energy is zero at $P=1.53\times 10^{9}\ \mathrm{Pa}=15\,300\ \mathrm{bar}$

Diamonds are formed naturally at even higher pressures and higher temperatures.  But this is the approximate order of magnitude.

So at $25\ \mathrm{^\circ C}$, graphite is highly thermodynamically favored over diamond.  The actual presence of diamonds at $25\ \mathrm{^\circ C}$ must therefore be a matter of kinetics, rather than thermodynamics.