This isn't a completely satisfying heuristic argument - in particular, it only partially specifies what compression must look like - but it does at least (a) use the matrix-based definition of free convolution and free compression, and (b) avoids extensive computations.

From the matrix-based definition of free convolution we see that the operation $\boxplus$ is both commutative and associative, and also torsion-free.  This is consistent with the existence of a large number of (semigroup) homomorphisms from probability measures equipped with $\boxplus$ to the real line equipped with addition.  And indeed we have the [free cumulants][1] $\kappa_1(\mu), \kappa_2(\mu), \dots$ with
$$ \kappa_n(\mu \boxplus \nu) = \kappa_n(\mu) + \kappa_n(\nu).$$
(These are secretly encoding the $R$-transform, but we will not need to explicitly work with the $R$-transform here.)

Now one has to take the following facts on faith:

 1. The free cumulants $\kappa_n(\mu)$ uniquely determine the measure.
 2. We have one free cumulant for each homogeneity: if $\lambda_* \mu$ denotes the pushforward of $\mu$ by the dilation map $x \mapsto \lambda x$, then $\kappa_n(\lambda_* \mu) = \lambda^n \kappa_n(\mu)$.

Now we turn to compression.  From the matrix-based definition we see that compression is an endomorphism on probability measures with free convolution:

$$ (\mu \boxplus \nu)^{\boxplus k} = \mu^{\boxplus k} \boxplus \nu^{\boxplus k}.$$

Thus (by fact 1), compression $\mu \mapsto \mu^{\boxplus k}$ is (morally at least) a linear map from the tuple $(\kappa_n(\mu))_{n=1}^\infty$ of free cumulants of $\mu$ to the tuple $(\kappa_n(\mu^{\boxplus k}))_{n=1}^\infty$.  On the other hand, it is clear from the matrix-based definition that compression also preserves homogeneity:
$$ ( \lambda_* \mu )^{\boxplus k} = \lambda_* (\mu^{\boxplus k}).$$
Thus this linear map must completely decouple in the cumulants (which describe the "isotypic components" of the homgeneity symmetry), and we must have a proportionality relationship of the form
$$\kappa_n(\mu^{\boxplus k}) = C_{n,k} \kappa_n(\mu)$$
for some constants $C_{n,k}$.  The matrix-based definition also gives the semigroup law
$$ (\mu^{\boxplus k})^{\boxplus l} = \mu^{\boxplus kl}$$
so this suggests (heuristically at least) that the relationship is of power type in $k$:
$$\kappa_n(\mu^{\boxplus k}) = k^{C_n} \kappa_n(\mu).$$
The one thing I don't have a slick explanation for is why $C_n=1$.  For $n=2$ one can see this just by plugging in the semicircular law which corresponds to GUE, using the matrix-based fact that the compression of GUE is just a rescaled version of GUE.  Unfortunately for higher $n$ the cumulants of the semicircular law vanish.  Perhaps there is another test distribution with non-vanishing cumulants for which the compression can be easily computed.  One could in principle work things out using an asymptotic analysis as $k \to \infty$ but this level of calculation begins to reach the same level as the usual free probability computations.

  [1]: https://en.wikipedia.org/wiki/Cumulant#Free_cumulants