Why is the spectroscopic entropy of carbon monooxide higher than that of nitrogen?
One contribution is the number of identical conformations of a molecule that can be generated by symmetry operations, which is described by a symmetry number $\sigma$. In the high temperature limit
$$S_\mathrm{rot} /R = 1 + \ln\left( \frac{8\pi^2IkT}{h^2\sigma} \right) $$
A larger symmetry number decreases the entropy. This number is 2 for nitrogen and 1 for carbon monoxide. That alone can account for $\log(2)=  0.69 \times R$. The larger symmetry number for nitrogen decreases the entropy by this amount.
In addition the moments of inertia (inversely proportional to the rotational temperature) of the two molecules are not really identical (it is ~5% larger for $\ce{CO}$), but the difference accounts for only ~$0.05 \times R$.