Why there are no $uuu$ and $ddd$ baryons with spin 1/2?

You are correct to point out that there's no symmetry that forbids a state with isospin 3/2 and spin 1/2; in the nomenclature, this is also called a $\Delta$ resonance. The Particle Data Group lists two such particles, with mass 1620 MeV and 1910 MeV. They exist, but they are heavier than the spin-3/2 $\Delta$ at 1232 MeV.

The reason why is isospin, although the exclusion principle is involved.

From the standpoint of the strong nuclear interaction, you can sometimes treat the proton and the neutron as two states of the same particle, the "nucleon." In quantum mechanics, a system with two internally available states usually tends to follow the same mathematical rules as a spinor with angular momentum ℏ/2; this is the case for the nucleon. So the strong interaction operator that distinguishes between is a "rotation" in "isotope space," or isospin.

Isospin is a good quantum number for the ground states and excited states of many light nuclei. In heavy nuclei, where the energy due to electrostatic repulsion starts to compete with the nuclear binding energy, the symmetry between proton and neutron is broken and you can't assign a definite isospin to a particular state.

In isotope space the pion is a three-state triplet, obeying the same algebra as a spin-one system in angular momentum space. You can think of the $\pi^+$ and $\pi^-$ as the isotopic raising and lowering operators on the proton and the neutron.

Similarly, a $\Delta$ is a strongly-interacting particle with total isospin 3/2. The $\Delta$ has four projections onto the charge axis, corresponding to the four charge states: $\Delta^{++}, \Delta^+, \Delta^0, \Delta^-$. Historically I believe the existence of the $\Delta^{++}$ with spin 3/2 was a lynchpin in the argument for the existence of quark color. The $\Delta^{++}(1232)$ has spin 3/2, so its spin wavefunction is symmetric under exchange; its isospin wavefunction, for the same reason, is symmetric under exchange; therefore there must be another degree of freedom with three states so that the quark wavefunction can be antisymmetric.

So why is a spin-1/2 $\Delta$ heavier than the lightest spin-3/2 $\Delta$? You can compare with the case of the deuteron. Nucleons don't have the color degree of freedom, so exchange symmetry — the exclusion principle — requires that a two-nucleon system with spin 0 must have isospin 1, and vice-versa. Isospin symmetry tells us that a proton-neutron pair with spin 0 should have roughly the same energy as a diproton or a dineutron. Since neither of those systems is bound, we expect to find the deuteron with isospin 0 and spin 1. Which it has. Apparently, in baryons and light nuclei, total isospin contributes more to the total energy of a system than does total angular momentum.

Correction

$\Delta(1620) 1/2^-$ is actually pretty well settled. (Thanks to rob.)

Original Answer

Actually, the Pauli exclusion principle can explain why there are no (uuu,ddd,sss) spin-1/2 ground states.

In baryons, quarks have four degree of freedom: orbital, spin, flavor, color. As you already know, the quarks' total wave functions should be anti-symmetric.

1. If we have uuu(or ddd, or sss) then the flavor part is symmetric;
2. Since we assume they are ground states,the orbital part is also symmetric;
3. For baryons(three-quark bound state), the color part is always anti-symmetric;

So we conclude that the spin part must be symmetric. It is the case for spin-3/2 states (baryon decuplet) not for spin-1/2 baryons (they have mixed symmetry). That's why uuu spin-1/2 ground state doesn't exist.

However, if the spin-1/2 uuu state is excited, then the orbital part of the wave function may also have mixed-symmetry. The orbital-spin can combine together (by group theory calculations) to be symmetric, the Pauli exclusion principle can't forbid this.

The only thing left is to find these state on experiments. This is not easy. As already pointed by user12262, there are only signs of the existence of $\Delta$ spin-1/2 states at present.