I've scourged the internet and I can't seem to find it anywhere although I've found the specific heat values for aluminum in both solid and liquid states. I'm trying to construct a heating curve and I need the specific heat capacity of aluminum in its gaseous state in order to calculate the amount of energy that would be necessary to bring aluminum from its liquid state to a gaseous state.
By "bring aluminum from its liquid state to a gaseous state", I assume you mean to a gaseous state at a temperature above aluminum's boiling point—since to merely convert it from a liquid to a gas, you only need the enthalpy of vaporization.
According to https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node18.html, most metallic vapors have molar heat capacities very close to those of ideal gases, i.e., $C_v = 3/2 R = 12.47$ J/(mol K) and $C_p = 5/2 R = 20.79$ J/(mol K). Thus you can obtain aluminum's specific heat capacity by dividing by the atomic mass, giving $C_v = 0.46$ J/(g K) and $C_p = 0.77$ J/(g K)
The reason for this is that metals don't form metallic bonds in the gaseous state (metallic bonding is a collective phenomenon that requires bulk metal), and are thus monatomic*. And the heat capacities of monatomic gases are simple, since they have only translational degrees of freedom.
[*As real gases, they may not be exactly monatomic, since they can still transiently associate via van der Waals forces. But treating them as monatomic is a good approximation.]
The exception to the above would occur when the temperature is sufficiently high to cause the metal to ionize, in which case you would have a higher heat capacity because you would now have two (or more) particles for every ionized aluminum atom. But since aluminum's first ionization energy is 5.985 eV/atom = $577$ kJ/mol, and $RT$ at $3000$ K is $25$ kJ/mol, I don't expect there to be significant ionization close to its $2792$ K boiling point. [Also, some might argue that, once it starts ionizing, it is no longer a gas, but a gas-plasma mixture.]