How do steric effects affect inductive effect in compounds with tetravalent nitrogen?
Since you are asking for a theoretical explanation, I believe a theoretical calculation might be helpful. It is possible to calculate the partial charges on atoms using electronic structure programs. I have calculated the partial charges at $\text{B3LYP/pcseg-1//GFN2-xTB}$ level (with GAMESS and xTB).
$\ce{Me-\overset{+}{N}(^iPr)3}$
We only need to look at the partial charges on the Me group to know the electron-withdrawing capacity of $\ce{-N+(^iPr)_3}$ group.
\begin{array}{cc} \hline
\mathrm{atom} &amp; \mathrm{partial\;charge\;(e)} \\ \hline
\ce{C} &amp; -0.4367\\
\ce{H} &amp; \;\;\, 0.1954\\
\ce{H} &amp; \;\;\, 0.1957\\
\ce{H} &amp; \;\;\, 0.1966\\ \hline
\end{array}
Total charge on $\ce{Me} = 0.151$.
$\ce{Me-\overset{+}{N}(^tBu)3}$
Again, we only need to look at the Me group.
\begin{array}{cc} \hline
\mathrm{atom} &amp; \mathrm{partial\;charge\;(e)} \\ \hline
\ce{C} &amp; -0.3087\\
\ce{H} &amp; \;\;\, 0.1436\\
\ce{H} &amp; \;\;\, 0.1430\\
\ce{H} &amp; \;\;\, 0.1414\\ \hline
\end{array}
Total charge on $\ce{Me} = 0.1193$
So, the partial charges would indicate that $\ce{-N+(^iPr)3}$ is more electron withdrawing than $\ce{-N+(^tBu)3}$.
Explanation?
I am not sure I have a definitive explanation for this. But it's possible to look at the differences between $\ce{^iPr}$ and $\ce{^tBu}$—

The cone angle formed by the three R groups for $\ce{-N+R3}$ would be larger for $\ce{R=^tBu}$ obviously, due to the steric bulk of that group. When the angles are the ideal $109.5^\circ$ each bond can be assumed to form from $\mathrm{N\;sp^3}$ orbitals. When N is $\mathrm{sp^2}$, the three bonds would be $120^\circ$ to each other. So, when the cone angle increases, the amound of p-character in those three N-R bonds can be assumed to go down. Conversely the amount of p-character in the 4th bond goes up which would mean less electron withdrawing capacity (p-orbitals are further away from nucleus). So, $\ce{-N+(^tBu)3}$ should be weaker EWG.

As there is no empty p-orbitals on N, so there should be no hyperconjugative effects for the compounds drawn. However, if instead of $\ce{Me}$, there is some other group that has a $\pi$ system attached to N, then there is a possibility of hyperconjugation.

The $\ce{CH3}$ groups have a very weak electron donating effect, so $\ce{-N+(^tBu)3}$ will be slightly weaker EWG than $\ce{-N+(^iPr)3}$. *


Notice that points 1 and 3 predict the same trend, which matches with the calculated trend. However, it is difficult to tell which factor is the most important (and whether these factors actually mean anything).

* Curiously, the partial charge on $\ce{C}$ of the side chain goes up when its $\ce{H}$ is replaced by $\ce{CH3}$, which would suggest that $\ce{CH3}$ is actually electron-withdrawing. This does not match with the standard notion of methyl groups being electron donating. I am not really sure if there is any explanation for this discrepancy. But note that there have been some evidence that $\ce{-CH3}$ groups can act as EWG's in some circumstances. The linked paper specifically mentions in the abstract—&quot;That methyl groups attached to carbon atoms are electron donors must not be generally assumed.&quot;