We split into two cases $0 < p < 4$ and $p\ge 4$.

Case 1. $0 < p < 4$

Fact 1. Let $a, b, c, d \in \mathbb{R}$ with $abcd = 1$ and $0 < a + b + c + d < 4$. Then $$a^2 + b^2 + c^2 + d^2 - \frac12(a + b + c + d)^2 - 4 \ge 0$$ with equality when $(a, b, c, d) = (u, u, -1/u, -1/u)$ for all $1 < u < 1 + \sqrt{2}$. (The proof is given at the end.)

By Fact 1, we have, for given $p \in (0, 4)$, subject to $a, b, c, d \in \mathbb{R}$ and $abcd = 1$ and $a + b + c + d = p$, the minimum of $a^2 + b^2 + c^2 + d^2$ is $p^2/2 + 4$ (when $(a, b, c, d) = (u, u, -1/u, -1/u)$ where $u = p/4 + \sqrt{1 + p^2/16}$).

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Case 2. $p \ge 4$

First, we give the following auxiliary result (Fact 2). The proof is given at the end.

Fact 2. Let $a, b, c\in \mathbb{R}$ with $a < b < c$. Then there exists $x, y \in \mathbb{R}$ such that $a + b + c = 2x + y$, and $abc = x^2y$, and $a^2 + b^2 + c^2 > 2x^2 + y^2$.

Now we proceed. Let $p\ge 4$ be fixed. Consider the minimum of $a^2+b^2+c^2+d^2$ subject to $a, b, c, d\in \mathbb{R}$ and $abcd=1$ and $a+b+c+d=p$. We claim that at minimum, among $a, b, c, d$, there are at most two distinct values. Assume, for the sake of contradiction, that at minimum, it holds that WLOG $a < b < c$. By Fact 2, we have $a + b + c + d = x + x + y + d$ and $abcd = xxyd$. However, $a^2+b^2+c^2+d^2 > x^2 + x^2 + y^2 + d^2$ which contradicts the optimality of $a, b, c, d$. The claim is proved.

So, there are two cases: (i) $a = b = c$, and (ii) $a = b, c = d$.

(i) Let $a = b = c = u$. We have $d = 1/u^3$. We have $3u + 1/u^3 = p$. We have $a^2+b^2+c^2+d^2 = 3u^2 + 1/u^6$.

(ii) Let $a = b = u, c = d = v$. We have $u^2v^2 = 1$ and $2u+2v=p$. We have $a^2 + b^2 + c^2 + d^2 = 2u^2 + 2v^2 = p^2/2 - 4uv \ge p^2/2 - 4$ with equality if $u = p/4 + \sqrt{p^2/16 -1}, v = p/4 - \sqrt{p^2/16-1}$.

Then we can get the minimum.

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Proof of Fact 1.

WLOG, assume that $a \ge b \ge c \ge d$. Then we have $a \ge b > 0$ and $0 > c \ge d$.

From $abcd = 1$, we have $(a+b)^2(c+d)^2 \ge 16$ using $(u+v)^2 \ge 4uv$. Let $x := a + b > 0$ and $y := -(c+d) > 0$. Then $xy \ge 4$.

We have $$a^2 + b^2 + c^2 + d^2 - \frac12(a + b + c + d)^2 - 4$$ $$\ge \frac{(a+b)^2}{2} + \frac{(c+d)^2}{2} - \frac12(a + b + c + d)^2 - 4 = xy - 4 \ge 0.$$ The rest is easy.

We are done.

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Proof of Fact 2.

We have $y = a + b + c - 2x$. Then we have $abc = x^2(a+b+c-2x)$ or $$F(x) := 2x^3 - (a+b+c)x^2 + abc = 0. $$

We have $$a^2 + b^2 + c^2 - 2x^2 - y^2 = -6x^2 + 4(a+b+c)x - 2(ab + bc + ca).$$ Note that $-6x^2 + 4(a+b+c)x - 2(ab + bc + ca) > 0$ is equivalent to $x_1 < x < x_2$ where $$x_1 := \frac{a+b+c}3 - \frac13\sqrt{a^2+b^2+c^2-ab-bc-ca},$$ $$x_2 := \frac{a+b+c}3 + \frac13\sqrt{a^2+b^2+c^2-ab-bc-ca}.$$

We have $$F(x_1)F(x_2) = -\frac{1}{27}(b-c)^2(a-c)^2(a-b)^2 < 0.$$ Thus, $F(x) = 0$ has a real root in $(x_1, x_2)$.