How can the ground state electron configuration for a d3 complex correspond to an F term symbol?
Free ion

The ground-state term symbol is only $\mathrm{^4F}$ in the case of a free ion. If you take a closer look at the Tanabe-Sugano diagram, the $\mathrm{^4F}$ term only appears at the far left-hand side of the diagram, where $\Delta = 0$. $\Delta$ refers to the ligand-field splitting parameter, and $\Delta = 0$ indicates that there is no ligand field, i.e. a free ion.

The quantum number $L$ (total orbital angular momentum) of the ground state can be obtained by coupling the individual orbital angular momenta of the d electrons using a Clebsch-Gordan series. The way to do this is described in most physical chemistry textbooks under the Russell-Saunders coupling scheme. For example in Atkins 10th ed. it is in page 386 under the chapter on "Atomic structure and spectra".

(Note that the symbol $\Lambda$ is used for diatomic molecules, not atoms.)

$L$ is said to be a "good" quantum number, in that the operator $\hat{L}^2$ (almost - this neglects spin-orbit coupling) commutes with the Hamiltonian $\hat{H}$. Quantum mechanically this means that the $\hat{H}$ and $\hat{L}^2$ (almost) share a set of eigenstates, and therefore for every state of the Hamiltonian (which correspond to the electronic configurations we are familiar with) one can (almost) calculate the corresponding value of $L$.

$$\hat{L}^2|\psi\rangle = L(L+1)\hbar^2|\psi\rangle$$

Octahedral complex

The ground state term symbol of a $(\mathrm{t_{2g}})^3$ ion is $\mathrm{^4\!A_2}$, not $\mathrm{^4F}$!

The $\mathrm{t_{2g}}$ set comprises the $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$ and $\mathrm{d}_{xy}$ orbitals. These three d orbitals are what we call "real" spherical harmonics, which are linear combinations of the complex spherical harmonics which you have quoted. As such it is not possible to assign $m_l$ values like you have to the $\mathrm{t_{2g}}$ orbitals.

It's not correct to say that $\mathrm{d}_{xy}$ can have "either $m_l = +2$ or $-2$". That would mean that $\mathrm{d}_{xy}$, at any one point in time, is either equal to $Y_2^{+2}$ or equal to $Y_2^{-2}$, which makes no sense. It is not flip-flopping between the two spherical harmonics, it is its own thing: a linear combination of the two spherical harmonics, or a superposition, if you prefer that word. Furthermore, the spherical harmonics only have significance under spherical symmetry, where they act as simultaneous eigenstates of $\hat{H}$, $\hat{L}^2$, and $\hat{L}_z$. In octahedral symmetry, the spherical harmonics aren't of any significance at all and trying to "resolve" the $\mathrm{t_{2g}}$ orbitals into their components as spherical harmonics is physically unmeaningful (it helps with the maths, but that's all).

Under octahedral symmetry, the total orbital angular momentum $L$ is no longer a good quantum number (i.e. $\hat{L}$ no longer commutes with the Hamiltonian) and therefore the term symbol doesn't say anything about it!