Let me try to address some of your concerns with a specific example. You wrote a lot, so I'm not exactly sure this will answer the things you are most concerned about, but I think there are some useful considerations here.

The voltage due to a sphere of uniform charge distribution in its center is $(3/2)kQ/r$, where $k=1/4\pi\epsilon_0$, $Q$ is the total charge, and $r$ is the radius of the sphere. The potential on the outer edge of that sphere is $kQ/r$. They differ by 50%, not a small fraction, no mater how small the sphere is. These will be the "blobs" in my example (with a lowercase $r$).

If you wanted to calculate the energy it would take to assemble such a sphere from point charges taken from infinitely far away, it would not suffice to approximate the potential in the "blob" as constant. If you take the potential in the center as the potential throughout, you would be overestimating the energy by something on the order of 50% (idk probably more like 25%). I think this makes you correct in saying that the macroscopic formula would not correctly capture the energy it would take if you "reshaped the blobs."

And so if the energy it took to assemble your "blobs" was significant compared to the energy it took to put your blobs together, indeed this formula would be wrong (it would incorrectly estimate the energy it took to assemble the blobs): $$ \frac{1}{2}\sum_{\text{blob }i}\phi_{\text{macro}}(x_i)\iiint\rho_{\text{macro}}dV=\frac{1}{2}\sum_{\text{blob }i}\phi_{\text{macro}}(x_i)Q_i $$

However, if you took a trillion such spheres and assembled them into a bigger sphere of radius $R\gg r$, then the potential in the center of that bigger sphere would be approximately $(3/2) kNQ/R$, where $N$ is the number of smaller spheres. Consider a small sphere sitting near the middle of the big sphere. Since $R\propto N^{1/3}$, the potential it experiences is much bigger than $kQ/r$ (by a factor $\sim N^{2/3}$), so its self assembly energy is much smaller than the energy it took to move to the center. This is the sense in which your various forms of this work formula are correct - they are correct if self assembly energies of the blobs are negligible. And furthermore, this example shows that it's reasonably common that most of the energy does come from assembling the macroscopic charge distribution, not from making the microscopic charges that are a part of that.

You can also ensure this when you do your Lorentz averaging - make sure that your averaging spheres are big enough to make a continuous function, but small enough that the self-assembly energy of the all the spheres remains negligible.

Otherwise they might also be correct if you only wanted to calculate how much work it took to assemble the bigger charge distribution, assuming the smaller spheres came pre-assembled. This is often the framework in which we calculate the energy it took to assemble a group of point charges. The energy it takes to make a point charge is infinite, so only changes in energy are even worth attempting to calculate.

Let me try to address some of your concerns with a specific example. You wrote a lot, so I'm not exactly sure this will answer the things you are most concerned about, but I think there are some useful considerations here.

The voltage due to a sphere of uniform charge distribution in its center is $(3/2)kQ/r$, where $k=1/4\pi\epsilon_0$, $Q$ is the total charge, and $r$ is the radius of the sphere. The potential on the outer edge of that sphere is $kQ/r$. They differ by 50%, not a small fraction, no mater how small the sphere is. These will be the "blobs" in my example (with a lowercase $r$).

If you wanted to calculate the energy it would take to assemble such a sphere from point charges taken from infinitely far away, it would not suffice to approximate the potential in the "blob" as constant. If you take the potential in the center as the potential throughout, you would be overestimating the energy by something on the order of 50% (idk probably more like 25%). I think this makes you correct in saying that the macroscopic formula would not correctly capture the energy it would take if you "reshaped the blobs."

And so if the energy it took to assemble your "blobs" was significant compared to the energy it took to put your blobs together, indeed this formula would be wrong (it would incorrectly estimate the energy it took to assemble the blobs): $$ \frac{1}{2}\sum_{\text{blob }i}\phi_{\text{macro}}(x_i)\iiint\rho_{\text{macro}}dV=\frac{1}{2}\sum_{\text{blob }i}\phi_{\text{macro}}(x_i)Q_i $$

However, if you took a trillion such spheres and assembled them into a bigger sphere of radius $R\gg r$, then the potential in the center of that bigger sphere would be approximately $(3/2) kNQ/R$, where $N$ is the number of smaller spheres. Consider a small sphere sitting near the middle of the big sphere. Since $R\propto N^{1/3}$, the potential it experiences is much bigger than $kQ/r$ (by a factor $\sim N^{2/3}$), so its self assembly energy is much smaller than the energy it took to move to the center. This is the sense in which your various forms of this work formula are correct - they are correct if self assembly energies of the blobs are negligible. And furthermore, this example shows that it's reasonably common that most of the energy does come from assembling the macroscopic charge distribution, not from making the microscopic charges that are a part of that.

You can also ensure this when you do your Lorentz averaging - make sure that your averaging spheres are big enough to make a continuous function, but small enough that the self-assembly energy of the all the blobs remains negligible.

Otherwise they might also be correct if you only wanted to calculate how much work it took to assemble the bigger charge distribution, assuming the smaller spheres came pre-assembled. This is often the framework in which we calculate the energy it took to assemble a group of point charges. The energy it takes to make a point charge is infinite, so only changes in energy are even worth attempting to calculate.

Let me try to address some of your concerns with a specific example. You wrote a lot, so I'm not exactly sure this will answer the things you are most concerned about, but I think there are some useful considerations here.

The voltage due to a sphere of uniform charge distribution in its center is $(3/2)kQ/r$, where $k=1/4\pi\epsilon_0$, $Q$ is the total charge, and $r$ is the radius of the sphere. The potential on the outer edge of that sphere is $kQ/r$. They differ by 50%, not a small fraction, no mater how small the sphere is. These will be the "blobs" in my example (with a lowercase $r$).

If you wanted to calculate the energy it would take to assemble such a sphere from point charges taken from infinitely far away, it would not suffice to approximate the potential in the "blob" as constant. If you take the potential in the center as the potential throughout, you would be overestimating the energy by something on the order of 50% (idk probably more like 25%). I think this makes you correct in saying that the macroscopic formula would not correctly capture the energy it would take if you "reshaped the blobs."

And so if the energy it took to assemble your "blobs" was significant compared to the energy it took to put your blobs together, indeed this formula would be wrong (it would incorrectly estimate the energy it took to assemble the blobs): $$ \frac{1}{2}\sum_{\text{blob }i}\phi_{\text{macro}}(x_i)\iiint\rho_{\text{macro}}dV=\frac{1}{2}\sum_{\text{blob }i}\phi_{\text{macro}}(x_i)Q_i $$

However, if you took a trillion such spheres and assembled them into a bigger sphere of radius $R\gg r$, then the potential in the center of that bigger sphere would be approximately $(3/2) kNQ/R$, where $N$ is the number of smaller spheres. Consider a small sphere sitting near the middle of the big sphere. Since $R\propto N^{1/3}$, the potential it experiences is much bigger than $kQ/r$ (by a factor $\sim N^{2/3}$), so its self assembly energy is much smaller than the energy it took to move to the center. This is the sense in which your various forms of this work formula are correct - they are correct if self assembly energies of the blobs are negligible. And furthermore, this example shows that it's reasonably common that most of the energy does come from assembling the macroscopic charge distribution, not from making the microscopic charges that are a part of that.

Otherwise they might also be correct if you only wanted to calculate how much work it took to assemble the bigger charge distribution, assuming the smaller spheres came pre-assembled. This is often the framework in which we calculate the energy it took to assemble a group of point charges. The energy it takes to make a point charge is infinite, so only changes in energy are even worth attempting to calculate.

Let me try to address some of your concerns with a specific example. You wrote a lot, so I'm not exactly sure this will answer the things you are most concerned about, but I think there are some useful considerations here.

The voltage due to a sphere of uniform charge distribution in its center is $(3/2)kQ/r$, where $k=1/4\pi\epsilon_0$, $Q$ is the total charge, and $r$ is the radius of the sphere. The potential on the outer edge of that sphere is $kQ/r$. They differ by 50%, not a small fraction, no mater how small the sphere is. These will be the "blobs" in my example (with a lowercase $r$).

If you wanted to calculate the energy it would take to assemble such a sphere from point charges taken from infinitely far away, it would not suffice to approximate the potential in the "blob" as constant. If you take the potential in the center as the potential throughout, you would be overestimating the energy by something on the order of 50% (idk probably more like 25%). I think this makes you correct in saying that the macroscopic formula would not correctly capture the energy it would take if you "reshaped the blobs."

And so if the energy it took to assemble your "blobs" was significant compared to the energy it took to put your blobs together, indeed this formula would be wrong (it would incorrectly estimate the energy it took to assemble the blobs): $$ \frac{1}{2}\sum_{\text{blob }i}\phi_{\text{macro}}(x_i)\iiint\rho_{\text{macro}}dV=\frac{1}{2}\sum_{\text{blob }i}\phi_{\text{macro}}(x_i)Q_i $$

However, if you took a trillion such spheres and assembled them into a bigger sphere of radius $R\gg r$, then the potential in the center of that bigger sphere would be approximately $(3/2) kNQ/R$, where $N$ is the number of smaller spheres. Consider a small sphere sitting near the middle of the big sphere. Since $R\propto N^{1/3}$, the potential it experiences is much bigger than $kQ/r$ (by a factor $\sim N^{2/3}$), so its self assembly energy is much smaller than the energy it took to move to the center. This is the sense in which your various forms of this work formula are correct - they are correct if self assembly energies of the blobs are negligible. And furthermore, this example shows that it's reasonably common that most of the energy does come from assembling the macroscopic charge distribution, not from making the microscopic charges that are a part of that.

You can also ensure this when you do your Lorentz averaging - make sure that your averaging spheres are big enough to make a continuous function, but small enough that the self-assembly energy of the all the spheres remains negligible.

Otherwise they might also be correct if you only wanted to calculate how much work it took to assemble the bigger charge distribution, assuming the smaller spheres came pre-assembled. This is often the framework in which we calculate the energy it took to assemble a group of point charges. The energy it takes to make a point charge is infinite, so only changes in energy are even worth attempting to calculate.