Consider all polynomials of a complex variable, $P(z)=4z^4+az^3+bz^2+cz+d$, where $a,b,c,$ and $d$ are integers, $0\le d\le c\le b\le a\le 4$, and the polynomial has a zero $z_0$ with $|z_0|=1.$ What is the sum of all values $P(1)$ over all the polynomials with these properties?
First, we consider the cases where $z_0 = 1$ and $z_0 = -1.$  Note that
\[P(1) = 4 + a + b + c + d \ge 4,\]so $z = 1$ cannot be a root of $P(z).$

If $z = -1$ is a root of $P(z),$ then
\[P(-1) = 4 - a + b - c + d = (4 - a) + (b - c) + d = 0.\]But $4 - a \ge 0,$ $b - c \ge 0,$ and $d \ge 0,$ so we must have $a = 4,$ $b = c,$ and $d = 0.$  Conversely, if $a = 4,$ $b = c,$ and $d = 0,$ then
\[P(-1) = 4 - a + b - c + d = (4 - a) + (b - c) + d = 0,\]so $z = -1$ is a root.  In this case,
\[P(1) = 4 + a + b + c + d = 4 + 4 + b + b = 8 + 2b.\]The sum of all possible values of $P(1)$ are then
\[\sum_{b = 0}^4 (8 + 2b) = 60.\]Having exhausted the cases where $z_0 = 1$ or $z_0 = -1,$ we can then assume that $z_0$ is not real.  Let $z_0 = x_0 + iy_0,$ where $x_0$ and $y_0$ are real numbers, $y_0 \neq 0.$  Since $|z_0| = 1,$ $x_0^2 + y_0^2 = 1.$  And since the coefficients of $P(z)$ are real, $x_0 - iy_0$ must also be a root, so
\[(z - x_0 - iy_0)(z - x_0 + iy_0) = z^2 - 2x_0z + x_0^2 + y_0^2 = z^2 - 2x_0 z + 1\]must be a factor of $P(z).$  Then
\[P(z) = (z^2 - 2x_0 z + 1)(4z^2 + pz + d)\]for some real number $p.$  Expanding, we get
\[P(z) = 4z^4 + (p - 8x_0) z^3 + (d - 2px_0 + 4) z^2 + (p - 8x_0) z + d.\]Comparing coefficients, we get
\begin{align*}
p - 8x_0 &= a, \\
d - 2px_0 + 4 &= b, \\
p - 2dx_0 &= c.
\end{align*}Subtracting the first and third equations, we get $2dx_0 - 8x_0 = a - c,$ so
\[2(d - 4) x_0 = a - c. \quad (*)\]If $d = 4,$ then $a = c.$  In fact, the chain $d \le c \le b \le a \le 4$ forces $a = b = c = d = 4,$ so
\[P(z) = 4z^4 + 4z^3 + 4z^2 + 4z + 4 = 4(z^4 + z^3 + z^2 + z + 1) = 0.\]If $z^4 + z^3 + z^2 + z + 1 = 0,$ then
\[(z - 1)(z^4 + z^3 + z^2 + z + 1) = 0,\]which becomes $z^5 - 1 = 0.$  Then $z^5 = 1,$ so $|z^5| = 1.$  Hence, $|z|^5 = 1,$ so $|z| = 1.$  This confirms that all the roots of $z^4 + z^3 + z^2 + z + 1$ have magnitude 1, and $P(1) = 20.$

Otherwise, we can assume that $d \neq 4.$  Then from equation $(*),$
\[2x_0 = \frac{a - c}{d - 4}.\]Multiplying the equation $p - 8x_0 = a$ by $d,$ we get
\[dp - 8dx_0 = ad.\]Multiplying the equation $p - 2dx_0 = c$ by 4, we get
\[4p - 8dx_0 = 4c.\]Subtracting these equations, we get $dp - 4p = ad - 4c,$ so
\[p = \frac{ad - 4c}{d - 4}.\]Let
\[k = 2px_0 = 2x_0 \cdot p = \frac{a - c}{d - 4} \cdot \frac{ad - 4c}{d - 4} = \frac{(a - c)(ad - 4c)}{(d - 4)^2}.\]Then from the equation $d - 2px_0 + 4 = b,$ $k = d - b + 4.$  Since $b \le 4,$ $k \ge 0.$  We then divide into the cases where $a = c$ and $a > c.$

Case 1: $a=c$.

In this case, $k=0$ and $b=d+4$, so $a=b=c=4$ and $d=0$.  We have already covered these possibilities when we looked at the case where $z = -1$ was a root of $P(z).$

Case 2: $a>c\geq 0$.

Since $k\geq 0$, we have $ad-4c\geq 0,$ or $ad \ge 4c$.  However, $ad \leq 4c$, so $ad = 4c$.  For this to hold, we must have $c = d.$  Then we obtain $k=0$ again. In this case, $b=d+4$, so $a=b=4$ and $c=d=0,$ and
\[P(z) = 4z^4 + 4z^3 + 4z^2 = 4z^2 (z^2 + z + 1).\]The roots of $z^2 + z + 1 = 0$ are $z = -\frac{1}{2} \pm \frac{\sqrt{3}}{2} i,$ which have magnitude 1, and $P(1) = 12.$

Therefore, the desired sum is $60 + 20 + 12 = \boxed{92}$.