Why does the emission wavelength of sodium appear to be an exception despite the trend for ionisation enthalpies among alkali metals?
On the NIST website you can look up the transitions corresponding to the wavelengths you list:

$$
\begin{array}{l c c c c}
\hline
\text{Element} &amp; λ/\pu{nm}&amp; |i\rangle &amp; |f\rangle &amp; \text{I.E./cm$^{-1}$}\\
\hline
\ce{Li} &amp; 670.8 &amp; 1s^22s &amp; 1s^22p &amp; 43\,487\\
\ce{Na} &amp; 589.2 &amp; 2p^63s &amp; 2p^63p &amp; 41\,449\\
\ce{K} &amp; 766.5 &amp; 3p^64s &amp; 3p^64p &amp; 35\,010\\
\ce{Rb} &amp; 780.0 &amp; 4p^65s &amp; 4p^65p &amp; 33\,691 \\
\ce{Cs} &amp; 455.5 &amp; 5p^66s &amp; 5p^67p &amp; 31\,406\\
\hline
\end{array}
$$

You see that the transition in Cs that you give in the table corresponds to a $7p\leftarrow 6s$ transition and that for consistency you should compare it with a $6p\leftarrow 6s$ transition which has a wavelength around 860 nm (no $d$ electrons involved). 

The term values of the electronic states of multi-electron atoms are relatively well described by Rydberg's formula

$$
\tilde{\nu}_{n,\ell}=\text{I.E.} - \frac{\mathcal{R}_M}{(n-\delta_\ell)^2},
$$
with $\mathcal{R}_M$ the mass-corrected Rydberg constant and where the quantum defect $\delta_\ell$ depends on the orbital angular momentum of the electron $\ell$ and on the properties of the ion core. The transition $n'p\leftarrow n''s$ can therefore be approximated by 
$$
\tilde{\nu}_{n',\ell=1}-\tilde{\nu}_{n'',\ell=0}=\frac{\mathcal{R}_M}{(n''-\delta_{\ell=0})}-\frac{\mathcal{R}_M}{(n'-\delta_{\ell=1})},
$$
and thus depend on the quantum defects for the $s$ and $p$ states. The quantum defects depend on the scattering properties of the Rydberg electron with the ion core and it is difficult to predict their values for different elements. It is therefore difficult to make quantitative predictions. Also for the heavier elements (higher nuclear charge $Z$), relativistic effects become more important, which influences the electronic structure.

Note that these lowest states are not really Rydberg states and the Rydberg formula is a bad approximation. For high values of $n$ the Rydberg formula gets much more accurate. In fact, it allows for a very accurate determination of the ionization energies of these elements.