Why isn't the orbital angular momentum also considered while calculating the magnetic moments 3d transition elements?
Simplistically speaking orbital angular momentum is present when some conditions are satisfied:


A set of orbitals are degenerate;
These orbitals can be "interconverted" by rotation about a certain axis;
The set of orbitals is not empty, half-filled, or fully filled.


Physically speaking, the "rotation" of an electron from one orbital to the next generates orbital angular momentum. The rotation must not have an energy cost, which necessitates degeneracy. It also means that the set of orbitals must have at least one electron, as well as one empty space (a so-called hole) for the electron to "rotate" into. The hole must have the same spin as the electron, which means that a half-filled subshell doesn't allow for orbital angular momentum (since all the electrons have parallel spins).

This is most easily illustrated with the three 2p orbitals in an atom. Rotation of a $\mathrm{p}_x$ orbital by 90 degrees around the $y$-axis, for example, turns it into a $\mathrm{p}_z$ orbital.

Any atom with a $\mathrm{p}^1$, $\mathrm{p}^2$, $\mathrm{p}^4$, or $\mathrm{p}^5$ configuration therefore possesses nonzero angular momentum. This is also reflected in their term symbols where $L = 1$. This is said to be "one unit of angular momentum". An atom with a half-filled shell (for example, $\ce{N}$)  or a closed-shell $\mathrm{p}^6$ configuration (for example, $\ce{Ne}$) does not have any orbital angular momentum.

The five 3d orbitals in an atom also satisfy these requirements. It is less intuitive to see this as compared to the 2p orbitals, because we have taken specific linear combinations of the complex 3d orbitals and converted them into real orbitals. Nevertheless one can use quantum mechanics to prove that the 3d orbitals possess angular momentum. The maths can be found in Figgis and Hitchman's Ligand Field Theory and Its Applications (it is very early in the book, but I do not have it with me now and cannot remember offhand where it is). As such, for free 3d metal atoms or ions, the orbital angular momentum has to be incorporated into the calculation of the magnetic moment. Exceptions are, of course, $\mathrm{d^0}$, $\mathrm{d^5}$, and $\mathrm{d^{10}}$ ions (e.g. $\ce{Sc^3+}$, $\ce{Mn^2+}$, and $\ce{Zn^2+}$).



That is applicable for free ions. However, 3d metals do not commonly exist as free ions, but exist in coordination complexes with a ligand field. For simplicity we'll consider a perfect octahedral ligand field, which causes d orbitals to split into a $\mathrm{t_{2g}}$ ($\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$) and $\mathrm{e_{g}}$ ($\mathrm{d}_{z^2}$, $\mathrm{d}_{x^2-y^2}$) set. This means that the five d orbitals are no longer all degenerate, breaking the first condition stipulated above.

At a simplistic level, many textbooks will claim that this leads to a "quenching" of orbital angular momentum, which just means that there is no longer any orbital angular momentum, and as such, the spin-only formula applies. [Note that there are some second-order effects (quantum mechanical mixing of excited states into the ground state) that can lead to deviations from the spin-only formula.]

This is true to a certain extent. However, the five 3d orbitals are not split into individually nondegenerate orbitals; the three $\mathrm{t_{2g}}$ orbitals, for example, are still degenerate. On top of that, one can verify pictorially that it is possible to rotate the $\mathrm{t_{2g}}$ orbitals into one another. In this aspect the $\mathrm{t_{2g}}$ orbitals behave quite similarly to the three $\mathrm{p}$ orbitals in a free atom; this is called the "$\mathrm{t_2}$-$\mathrm{p}$ isomorphism". As such, the $\mathrm{t_{2g}}$ orbitals still have orbital angular momentum associated with them.

Any complex with unsymmetrical occupancy of the $\mathrm{t_{2g}}$ orbitals (i.e. $\mathrm{t_{2g}}^1$, $\mathrm{t_{2g}}^2$, $\mathrm{t_{2g}}^4$, $\mathrm{t_{2g}}^5$) therefore possesses orbital angular momentum and exhibit great deviations from the spin-only formula. These complexes all have a $\mathrm{T_{1g}}$ or $\mathrm{T_{2g}}$ ground state term symbol, so it is commonly said that "$\mathrm{T}$ terms possess one unit of orbital angular momentum". Spin-orbit coupling has to be invoked to obtain a better agreement between the calculated and measured magnetic moments.

As an example of how large the deviations can be, consider the compound $\ce{(NH4)2Co(SO4)2.2H2O}$. This features octahedral Co(II) which adopts a high-spin configuration $(\mathrm{t_{2g}})^5(\mathrm{e_g})^2$. From the spin-only formula, we would expect $\mu_\mathrm{so} = \sqrt{3(3+2} = 3.87$. However, the unsymmetrical population of the $\mathrm{t_{2g}}$ orbitals leads to the generation of orbital angular momentum, and at $300~\mathrm{K}$ the measured effective magnetic moment is actually $5.1$. (Data from here.)



A more detailed discussion can be found in any of the standard inorganic chemistry textbooks. Figgis and Hitchman's book (mentioned earlier) is more advanced and has a very comprehensive treatment of the subject.