Decided to practice my knowledge of representation theory by constructing the weight diagram for the representation $(2 \, 1)$ of $SU(3)$. This is apparently the $\mathbf{15}$, but when I use what I recall of the method to construct the weights at lower levels using the simple roots, I arrive at dimension 11.
Here's the diagram, where the simple roots are $\alpha_1 = (2 ~ \bar{1})$ and $\alpha_2 = (\bar{1} ~ 2)$, and an overbar denotes a minus sign:
$ \large\qquad {(2~1)} \\ \quad {\alpha_1} \nearrow \quad \nwarrow \alpha_2\\ \large(0 ~ 2)\quad\quad(3~\bar{1})\\ ~\alpha_1\uparrow \qquad \quad \uparrow\alpha_1 \\ \large(\bar{2} ~ 3) \quad\quad (1~0)\\ \quad \alpha_2 \nwarrow \quad\nearrow \alpha_1 \\ \large~~~\qquad(\bar{1} ~ 1)\\ \quad\alpha_1\nearrow \quad \nwarrow \alpha_2\\ \large(\bar{3} ~~ 2) \quad \quad (0 ~ \bar{1})\\ ~\alpha_2\uparrow \qquad \uparrow \alpha_2\\ \large(\bar{2} ~ 0)\quad \quad (1 ~ \bar{3})\\ \quad \alpha_2\nwarrow \quad\nearrow\alpha_1 \\ \large~~~~\quad(\bar1 ~ \bar{2}) $
I'm able to correctly get the $\mathbf{10}$ and $\mathbf{8}$ using what I recall.
Some of the weights are repeated. The weights $(0,1),(-1,1),(0,-1)$ are repeated twice. Somehow you're also missing the weight $(2,-2)$, which occurs once. This weight would "sit" below $(-1,1)$ but above $(-3,2)$ and $(0,-1)$. With these adjustments you now sum up to $15$.
I'm pretty sure I have it right as I have a code which uses the Freudenthal formula to generate weights and their multiplicities, and I get exactly the same tower as you get (in reverse order as I start from the highest weight) except for the additional $(2,-2)$ weight located as described above. My code also produces the weight multiplicities using the same formula.