EM energy flux and heat distribution in a cylindrical conductor Consider a cylindrical conductor made of a single material (say, copper) with conductivity $\sigma$. Assume there is a current through the conductor and, to avoid complications, assume that the current is constant.
Question: Given that the vast majority of the EM (electromagnetic) energy flows near a conductor surface, why does the spatial temperature distribution ("heat-map") of the conductor (due to Joule heating) decrease radially outwards?
P.S. Note that I may simply be mistaken that the "heat-map" decreases radially outwards.
 A: For DC current, the current density $\vec J$ and the electric field $\vec E$ are both more or less uniform in the conductor. This means the Joule heating power density $\vec J \cdot \vec E$ is also approximately uniform in the conductor. However, the conductor cools from its surface, so the temperature is higher near the conductor's axis.
There are some second-order effects that cause $\vec J$ to fail to be exactly uniform. Most conductors have a positive temperature coefficient of resistance, which means the conductivity is lower around the conductor's axis (where the temperature is high), so the current density is lower here. Another effect is the usually minute Hall effect caused by the conductor's own magnetic field, resulting in some radial carrier redistribution and a corresponding radial variation in conductivity. But normally the temperature will still be higher near the center.
The energy density doesn't directly influence where the heat is generated. Note that if you regard the energy as being stored in the fields, the energy density is non-zero even outside the conductor, but this doesn't mean heat is generated here. You would see the effect of Joule heating on the Poynting vector in the form of a radially inward component as in the figure below, representing the flow of electromagnetic energy into the conductor where it is converted to heat.

I. Galili and E. Goihbarg, "Energy transfer in electrical circuits: A qualitative account", Am. J. Phys. 73, 141 (2005)
As for your question about why $\vec E \ne 0$ in the conductor, in general $\vec E = 0$ in a conductor is only valid in electrostatics where there is no current flow. A non-zero $\vec E$ is needed to push charges around and create current (except in superconductors).
A: Let’s choose a cylindrical coordinate system $\rho,\theta,z$ along the axis of the wire.  We’ll assume that the wire is the the positive conductor in some electric circuit, and therefore has a small nonzero net charge; see e.g. the illustration at this other answer.
Model the electric field as
\begin{align}
\vec E_\text{inside} &= \frac1\sigma \vec J = E_\text{in}\hat z
\\
\vec E_\text{outside} &= E_\text{out} \frac R\rho \hat \rho
\end{align}
where $\vec J$ is the current density,
$\sigma$ is the conductivity of the wire,
$R$ is the wire’s radius, and the field magnitudes $E_\text{in}$ and $E_\text{out}$ are the solutions to freshman homework problems.
Actually, we can see that this approximation is not strictly allowed.  A surface charge, such as on the edge of our wire, introduces a discontinuity in the field normal the surface, the $\hat \rho$ direction.  However the field components parallel to the surface, in the $\hat\theta$ and $\hat z$ directions, must be continuous.  So our outside field must actually be
$$
\vec E_\text{outside} = E_\text{out}\frac R\rho \hat\rho + E_\text{in} \hat z
$$
Our assumption that the field outside the wire is radial is therefore equivalent to an assumption that the conductivity $\sigma$ of the wire is large.
The magnetic field at some distance $\rho$ from the axis is
$$
\vec B = \frac{\mu_0}{2\pi} \frac{I_\text{enclosed}}{\rho} \hat \theta
$$
Outside of the wire, $\rho > R$, the enclosed current is just the entire current, $I = \pi R^2 J$.  Inside the wire, only the inner part of the current contributes, and the field becomes
$$
\vec B_\text{inside} = \frac{\mu_0}{2\pi}\frac{\pi\rho^2 J}{\rho}\hat\theta \propto \rho\hat\theta
$$
We can find the direction of the Poynting vector $\vec S \propto \vec E \times \vec B$ everywhere.  Outside of the wire there is a component proportional to
$$\vec S_\text{downstream} \propto E_\text{out} \hat\rho \times \hat\theta = E_\text{out} \hat z$$
That is to say, the power transport outside of the wire is parallel to the direction of the current in the “supply” wire of the circuit.  The current-return wire will have a small net negative charge, $\vec E{}_\text{outside}^\text{return} \propto -\hat\rho$, and energy will flow antiparallel to the current — also towards the load.
There is also a radial power flow proportional to
$$
\vec S_\text{radial} \propto E_\text{in} |\vec B| (-\hat\rho)
$$
which is responsible for heating the wire.  To find how much energy is deposited per unit volume, let’s consider a cylindrical shell with radius $r$, thickness $d\rho$, length $L$, and volume $2\pi rL\,d\rho$.
The power deposited in this shell is given the difference between the power entering from its outside and the power exiting towards its inside:
\begin{align}
P &=
\int_\text{outer} \vec S \cdot d\vec A
-
\int_\text{inner} \vec S \cdot d\vec A
\\
&= S_\text{outer} A_\text{outer} - S_\text{inner} A_\text{inner}
\end{align}
Outside of the wire, we have $S \propto B \propto r^{-1}$, and area $A \propto r$, so no energy is deposited in the current-free space outside of the wire.  However, inside the wire we have
$$
\vec S \cdot d\vec A \propto BA \propto r^2
$$
The power in a shell with thickness $d\rho$ therefore goes like
\begin{align}
B_\text{outer} A_\text{outer}
-
B_\text{inner} A_\text{inner}
& \propto
(r + d\rho)^2 - r^2 \approx 2r\, d\rho
\end{align}
Since the power per unit volume and the volume itself are both proportional to $r\,d\rho$, we have uniform power deposition in the metal.
You ask why the wire is hotter at the center than at the edges.  This is a heat-transport problem: the heat can only leave the wire through its surface, so the surface must cool before the interior.
