What restricts the value of weak hypercharge from being 5/3? The quarks and leptons chiral states have hypercharges of $0,\pm 1/3,\pm 2/3, \pm 3/3, \pm 4/3, \pm 6/3$. The value of $\pm 5/3$ is prominently missing. Is there a theoretical principle which picks out these values?
In looking for papers on arXiv with hypercharge 5/3, I've found only a handful of papers. One is a preprint from December 2010 (See page 16 "KK (Kaluza-Klein) towers") : 
Giuliano Panico, Mahmoud Safari, Marco Serone, (2010) Simple and Realistic Composite Higgs Models in Flat Extra Dimensions
http://arxiv.org/abs/1012.2875
And then they come up as a vector diquark here and in a few similar papers:
Journal of High Energy Physics 2002, 024, (2002), Engin Arik, Serkant A. Çetin, Orhan Çakir, and Saleh Sultansoy, A Search for Vector Diquarks at the CERN LHC
http://arxiv.org/abs/hep-ph/0109011v3
So is there a theoretical principle that picks out which weak hypercharge values might be observed in nature?
 A: Hypercharge assignments in the Standard Model (that is with just the standard leptons and quarks) are determined up to a twofold ambiguity by demanding cancellation of all anomalies. Here's a reference:J.A. Minahan, Pierre Ramond, (Florida U.) , R.C. Warner, (Melbourne U.) . UFIFT-HEP-89-16, Jun 1989. 6pp. 
Published in Phys.Rev.D41:715,1990.
A: If you consider particular enough models, the assignment of the allowed hypercharges may be easily calculated. The models themselves are constrained by various additional conditions such as anomaly cancellation.
For example, take an $SO(10)$ Grand Unified Theory. In that theory, all the fermions arise from the chiral spinorial ${\bf 16}$ representations of $SO(10)$ - or $spin(10)$, to be more accurate. It's the simplest case relevant for the Standard Model where it is easy to show that there are no anomalies carried by this ${\bf 16}$ representation: it boils down to the vanishing of 
$$\mbox{Tr}\left(\gamma_{ab} \{\gamma_{cd},\gamma_{ef}\}\right)$$
for all $a,b,c,d,e,f=1,\dots, 10$ which is trivial to check.
The weights of ${\bf 16}$ under $SO(10)$ are
$$ (\pm \frac12, \pm \frac12, \pm \frac12, \pm \frac12, \pm \frac12) $$
with an even number of pluses (and odd number of minuses) - or vice versa, for the antifermions. The hypercharge is the inner product of the weight of a state with the generator
$$ (+2,+2,+2,-3,-3)/3 $$
The first three entries are linked to the three colors; the last two entries are associated with the weak doublets. The sum of the coordinates above had to vanish because the hypercharge is also a generator of $SU(5)$ where all generators have a vanishing trace (the sum of 5 diagonal elements) because of the $S$.
Now, take all the inner products. They will be sums of $k_1=-3,-1,+1$, or $+3$ terms $(+2/2)\cdot (+1/2)=1/2$, and $k_2=-2, 0$, or $+2$ times $(+3/2)\cdot(+1/2)=3/4$. List all $4\times 3$ possible values of
$$ k_1\cdot \frac 12 + k_2\cdot \frac 34 $$
for the allowed values of $k_1,k_2$ and you will get exactly the 11+1 possibilities you listed (zero will appear twice because $SO(10)$ also predicts the right-handed neutrino).
This is the derivation of your list from a single representation, the ${\bf 16}$ of $SO(10)$. It's completely inherited by $SU(5)$ and the Standard Model - without the right-handed neutrino. (Some nonzero values of the hypercharge also appear multiply times - for several basis vectors - e.g. quarks appear thrice.)
Note that the maximum hypercharge I could have obtained from the inner products was 
$$(2+2+2+3+3)/3/2=6/3$$
I could have obtained any smaller multiple of $1/3$ because the terms $2/3$ and $3/3$ may conspire in various ways. However, I can't reduce $6/3$ just by $1/3$ because the minimum amounts I can subtract by changing a sign are $2/3$ or $3/3$.
A: In the hypothetical $SU(5)$ theory, the X and Y bosons have hypercharges of $\pm 5/3$. So do some of the components of the adjoint Higgs needed to break $SU(5)$ down to the Standard Model.
A: If you consider particular enough models, the assignment of the allowed hypercharges may be easily calculated. The models themselves are constrained by various additional conditions such as anomaly cancellation.
For example, take an SO(10) Grand Unified Theory. In that theory, all the fermions arise from the chiral spinorial 16 representations of SO(10) - or spin(10), to be more accurate. It's the simplest case relevant for the Standard Model where it is easy to show that there are no anomalies carried by this 16 representation: it boils down to the vanishing of
Tr(γab{γcd,γef})
for all a,b,c,d,e,f=1,…,10 which is trivial to check.
The weights of 16 under SO(10) are
(±12,±12,±12,±12,±12)
with an even number of pluses (and odd number of minuses) - or vice versa, for the antifermions. The hypercharge is the inner product of the weight of a state with the generator
(+2,+2,+2,−3,−3)/3
The first three entries are linked to the three colors; the last two entries are associated with the weak doublets. The sum of the coordinates above had to vanish because the hypercharge is also a generator of SU(5) where all generators have a vanishing trace (the sum of 5 diagonal elements) because of the S.
Now, take all the inner products. They will be sums of k1=−3,−1,+1, or +3 terms (+2/2)⋅(+1/2)=1/2, and k2=−2,0, or +2 times (+3/2)⋅(+1/2)=3/4. List all 4×3 possible values of
k1⋅12+k2⋅34
for the allowed values of k1,k2 and you will get exactly the 11+1 possibilities you listed (zero will appear twice because SO(10) also predicts the right-handed neutrino).
This is the derivation of your list from a single representation, the 16 of SO(10). It's completely inherited by SU(5) and the Standard Model - without the right-handed neutrino. (Some nonzero values of the hypercharge also appear multiply times - for several basis vectors - e.g. quarks appear thrice.)
Note that the maximum hypercharge I could have obtained from the inner products was
(2+2+2+3+3)/3/2=6/3
I could have obtained any smaller multiple of 1/3 because the terms 2/3 and 3/3 may conspire in various ways. However, I can't reduce 6/3 just by 1/3 because the minimum amounts I can subtract by changing a sign are 2/3 or 3/3.
