It may be useful to expand my comments. The question involves Lie type $C_m$ with $m \geq 2$. Without developing Lie group or algebraic group language, it's enough to work with a simple Lie algebra over $\mathbb{C}$ of this type. Using the standard numbering of vertices in the Dynkin diagram, let $E_i$ be the fundamental representation of highest weight $\varpi_i$ for $i= 1, \dots, m$. Here $E_1$ is the standard module of dimension $2m$. For the others, there are numerous classical references. There is a thorough discussion of the construction in Bourbaki Groupes et algebres de Lie (also in English translation), Chap. VIII, $\S13$, no. 3, (IV). In particular, the well-known dimension formula is made explicit:
$$\dim E_i = \binom{2m}{i} - \binom{2m}{i-2} \text{ for } i \geq 2$$.
Clearly $\dim E_1 > \dim E_2$. The claim is that $\dim E_2 \geq \dim E_j$
for all $j >2$. This should require an elementary combinatorial comparison, not involving any Lie theory, though it would be interesting to see a conceptual argument.
Granted this inequality, Weyl's dimension formula (as already noted) will complete the desired argument for $ E_2$ being the second smallest nontrivial irreducible representation. The formula involves a fraction, whose denominator can be ignored. The numerator is an integral polynomial in the highest weight, which obviously grows larger as the coordinates of that weight increase relative to the $\varpi_i$.
P.S. I don't want to leave the impression that I've written down a formal proof. It's only a proof-scheme, but should be fairly easy to complete using straightforward methods. For the comparison between fundamental and non-fundamental weights, you'd need to look at the root system $C_m$ (say at the end of Chapters IV-VI of Bourbaki): a rough comparison of how often $\alpha_1, \alpha_2$ occur in each positive root shows for instance how the Weyl dimension for $2\varpi_1$ exceeds the dimension for $\varpi_2$, etc. I don't recall seeing all of this written down anywhere, but if there is motivation to do so it should be elementary to complete.
$m=3$
gives (I hope)` dimensions 6, 14, 14, but after that the later ones grow faster. Presumably the answer to your question is yes, but it needs an argument. $\endgroup$