Assume the existence of a positive integer solution $(a, b, c, d, n)$ such that $a^2 + b^2 + c^2 + d^2 - 4\sqrt{abcd} = 7 \cdot 2^{2n-1}$, and take the one with the minimum value of $n$. Since $\sqrt{abcd}$ is rational, $abcd$ must be a square number. First, we establish the following two lemmas. Note that $a^2 + b^2 + c^2 + d^2 - 4\sqrt{abcd} = (a-b)^2 + (c-d)^2 + 2(\sqrt{ab} - \sqrt{cd})^2$.

Lemma 1. $a, b, c, d$ are all distinct from each other.
Proof. From the symmetry of $a, b, c, d$, it suffices to assume $a = b$ and derive a contradiction. Since $abcd = a^2cd$ is a square number, $cd$ must also be a square number. Therefore, both $M = c - d$ and $N = a - \sqrt{cd}$ are integers, and we have $M^2 + 2N^2 = 7 \cdot 2^{2n-1}$. Since the residues of square numbers divided by $7$ are $0, 1, 2,$ or $4$, the residues of $M^2$ and $2N^2$ when divided by $7$ must also be $0, 1, 2,$ or $4$. Since their sum is a multiple of $7$, both $M$ and $N$ must be multiples of $7$. Therefore, $M^2 + 2N^2$ is a multiple of $49$. However, $7 \cdot 2^{2n-1}$ is not a multiple of $49$, which is a contradiction. ■

Lemma 2. The product of any two numbers among $a, b, c, d$ is different from the product of the remaining two numbers.
Proof. From the symmetry of $a, b, c, d$, we only need to derive a contradiction by assuming $ab = cd$. Let $M = a - b$ and $N = c - d$, then $M$ and $N$ are integers, and $M^2 + N^2 = 7 \cdot 2^{2n-1}$ holds. Because the residues of square numbers divided by $7$ can only be $0, 1, 2,$ or $4$, as mentioned in Lemma 1, both $M$ and $N$ must be multiples of $7$. Therefore, $M^2 + N^2$ is a multiple of $49$, but $7 \cdot 2^{2n-1}$ is not, leading to a contradiction. ■

Consider cases based on the value of $n$.

(1) $n = 1$
By the symmetry of $a, b, c, d$ and Lemma 1, we can assume that $a < b < c < d$. From $(d-a)^2 + (c-b)^2 + 2(\sqrt{ad} - \sqrt{bc})^2 = 14$, we have $d-a \le \sqrt{14} < 4$, implying $a = b-1 = c-2 = d-3$. We also get $bc = a^2 + 3a + 2 = ad + 2$ and consequently $\sqrt{ad} + 2 - \sqrt{ad} = \sqrt{2}$, which leads to a contradiction since $a$ and $d$ are positive.

(2) $n > 1$
When $a, b, c, d$ are all even, $(\frac{a}{2}, \frac{b}{2}, \frac{c}{2}, \frac{d}{2}, n-1)$ satisfies the given equation, contradicting the minimality of $n$. Therefore, at least one of $a, b, c, d$ is odd. Since the residues of square numbers divided by $4$ are $0$ or $1$, and $a^2 + b^2 + c^2 + d^2 = 7 \cdot 2^{2n-1} + 4\sqrt{abcd}$ is divisible by $4$, all of $a, b, c, d$ are odd. Since $abcd$ is an odd square number, $abcd \equiv 1 \pmod{4}$, implying that there are an even number of elements among $a, b, c, d$ that leave a remainder of $3$ when divided by $4$. Thus, without loss of generality, we can assume $a \equiv b \pmod{4}$ and $c \equiv d \pmod{4}$. Let $g$ denote the greatest common divisor of $ab$ and $cd$. Then, $\frac{ab}{g}$ and $\frac{cd}{g}$ are coprime, and $\frac{ab}{g} \cdot \frac{cd}{g} = \frac{abcd}{g^2}$ is a square number. Hence, there exist positive integers $x$ and $y$ such that $ab = gx^2$ and $cd = gy^2$, and $(\sqrt{ab} - \sqrt{cd})^2 = g(x - y)^2$ holds. Since $a \equiv b \pmod{4}$, we have $ab \equiv 1 \pmod{4}$, and combined with $x^2 \equiv 1 \pmod{4}$, this yields $g \equiv 1 \pmod{4}$. By Lemma 1 and Lemma 2, we can find positive odd integers $p, q, r$ and non-negative integers $\alpha, \beta, \gamma$ satisfying $|a-b| = p \cdot 2^\alpha, |c-d| = q \cdot 2^\beta$, and $|x-y| = r \cdot 2^\gamma$. Remark that
$$\begin{aligned} 7 \cdot 2^{2n-1} &= a^2 + b^2 + c^2 + d^2 - 4\sqrt{abcd} \\ &= (a-b)^2 + (c-d)^2 + 2(\sqrt{ab} - \sqrt{cd})^2 \\ &= p^2 \cdot 2^{2\alpha} + q^2 \cdot 2^{2\beta} + gr^2 \cdot 2^{2\gamma+1} \end{aligned}$$
holds. $p^2 \equiv q^2 \equiv 1 \pmod{8}$, $gr^2 \equiv 1 \pmod{4}$ imply a contradiction by the following lemma.

Lemma 3. There exist no non-negative integers $\alpha, \beta, \gamma$, positive integers $n, s, t, u$ such that $s \equiv t \equiv 1 \pmod{8}$, $u \equiv 1 \pmod{4}$, and
$$s \cdot 2^{2\alpha} + t \cdot 2^{2\beta} + u \cdot 2^{2\gamma+1} = 7 \cdot 2^{2n-1}.$$
Proof. By the symmetry of $(\alpha, s)$ and $(\beta, t)$, we can assume $\alpha \le \beta$.

* case when $\alpha \le \gamma$
Since $s+t \cdot 2^{2(\beta-\alpha)} + u \cdot 2^{2(\gamma-\alpha)+1} = 7 \cdot 2^{2(n-\alpha)-1}$ is a positive integer, we must have $n-\alpha \ge 1$. Therefore, both $u \cdot 2^{2(\gamma-\alpha)+1}$ and $7 \cdot 2^{2(n-\alpha)-1}$ are even, implying that $s+t \cdot 2^{2(\beta-\alpha)}$ is even as well, and consequently, $\alpha = \beta$. Since $s+t \cdot 2^0$ cannot be divisible by $4$, we obtain either $\gamma-\alpha = 0$ or $n-\alpha = 1$. In the former case, $7 \cdot 2^{2(n-\alpha)-1} = s+t+2u \equiv 4 \pmod{8}$, leading to a contradiction. In the latter case, $u \cdot 2^{2(\gamma-\alpha)+1} = 14-s-t \equiv 4 \pmod{8}$, which also contradicts the conditions.

* case when $\gamma < \alpha$
Since $s \cdot 2^{2(\alpha-\gamma)} + t \cdot 2^{2(\beta-\gamma)} + 2u = 7 \cdot 2^{2(n-\gamma)-1}$ has a remainder of $2$ when divided by $4$, we must have $n-\gamma = 1$. In this case, $2^{2(\beta-\gamma)} < 14$, implying $\alpha-\gamma \le \beta-\gamma \le 1$, which leads to $\alpha-\gamma = \beta-\gamma = 1$. Therefore, we have $2s+2t+u = 7$, but $2s+2t+u \equiv 1 \pmod{4}$, which is a contradiction.

Therefore we have proved the lemma.

Therefore, there are no solutions $(a, b, c, d, n)$.