It will be easier if it is first proved that the determinant of a square matrix can be computed by expansion about any row or column; see this for the details.
I will use the $\rho$-notation which is defined in the above answer.
I will prove only the first theorem; the proof of the second one is very similar to that of the first one, which I leave as an exercise.
Let the proposition $P(k)$ be as follows:
For any $n \times n$ matrix (in which $n \geq k$) and for any $k$ distinct integers $j_1$, $j_2$, $\dots$, $j_k$ less than or equal to $n$ (in which $j_1 < j_2 < \dots < j_k$),
$$
\begin{aligned}
\det {(A)}
= \sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n}
{\det {\left(
A\binom{i_1, i_2, \dots, i_k}
{j_1, j_2, \dots, j_k}
\right)}}
(-1)^{i_1 + i_2 + \dots + i_k
+ j_1 + j_2 + \dots + j_k}
\det {(A({i_1,i_2,\dots,i_k}|{j_1,j_2,\dots,j_k}))}.
\end{aligned}
$$
We will prove that $P(k)$ is true for $k = 1$, $2$, $3$, $\dots$.
$P(1)$ is true, because the determinant of a square matrix can be computed by expansion about any column.
Suppose that $P(k-1)$ is true. Our goal is to prove that $P(k)$ is true.
Let $A$ be an $n \times n$ matrix, with $n \geq k$. Let $j_1$, $j_2$, $\dots$, $j_k$ be any $k$ distinct integers less than or equal to $n$, with $j_1 < j_2 < \dots < j_k$. Then
$$
\begin{aligned}
& \det {(A)}
\\
= {} &
\sum_{d_1 = 1}^{n}
{[A]_{d_1,j_1}
(-1)^{d_1 + j_1}
\det {(A(d_1|j_1))}}
\\
= {} &
\sum_{d_1 = 1}^{n}
{[A]_{d_1,j_1}
(-1)^{d_1 + j_1}}
\sum_{\substack{
1 \leq d_2 < \dots < d_k \leq n \\
d_1 \neq d_2,\dots,d_k
}}
{\det {\left(
A\binom{d_2, \dots, d_k}{j_2, \dots, j_k}
\right)}}
(-1)^{d_2 - \rho(d_2, d_1) + \dots
+ d_k - \rho(d_k, d_1)
+ j_2 + \dots + j_k - (k - 1)}
\det
{(A({d_1,d_2,\dots,d_k}|{j_1,j_2,\dots,j_k}))}
\\
= {} &
\sum_{\substack{
1 \leq d_1 \leq n \\
1 \leq d_2 < \dots < d_k \leq n \\
d_1 \neq d_2,\dots,d_k
}}
{[A]_{d_1,j_1}
(-1)^{d_1 + j_1}}
{\det {\left(
A\binom{d_2, \dots, d_k}{j_2, \dots, j_k}
\right)}}
(-1)^{d_2 - \rho(d_2, d_1) + \dots
+ d_k - \rho(d_k, d_1)
+ j_2 + \dots + j_k - (k - 1)}
\det
{(A({d_1,d_2,\dots,d_k}|{j_1,j_2,\dots,j_k}))}
\\
= {} &
\sum_{\substack{
1 \leq d_1 \leq n \\
1 \leq d_2 < \dots < d_k \leq n \\
d_1 \neq d_2,\dots,d_k
}}
(-1)^{
(d_1 + \dots + d_k) + (j_1 + \dots + j_k)
+ (\rho(d_1, d_2) + \dots + \rho(d_1, d_k))
- 2(k-1)
}
[A]_{d_1,j_1}
{\det {\left(
A\binom{d_2, \dots, d_k}{j_2, \dots, j_k}
\right)}}
\det
{(A({d_1,d_2,\dots,d_k}|{j_1,j_2,\dots,j_k}))}
\\
= {} &
\sum_{\substack{
1 \leq d_1 \leq n \\
1 \leq d_2 < \dots < d_k \leq n \\
d_1 \neq d_2,\dots,d_k
}}
(-1)^{
(d_1 + \dots + d_k) + (j_1 + \dots + j_k)
+ (\rho(d_1, d_2) + \dots + \rho(d_1, d_k))
}
[A]_{d_1,j_1}
{\det {\left(
A\binom{d_2, \dots, d_k}{j_2, \dots, j_k}
\right)}}
\det
{(A({d_1,d_2,\dots,d_k}|{j_1,j_2,\dots,j_k}))}
\\
= {} &
\sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n}
\sum_{\substack{
1 \leq s \leq k\\
d_1 = i_s \\
d_r = i_{r-1}, \ 2 \leq r \leq s \\
d_r = i_{r}, \ r > s \\
}}
(-1)^{
(d_1 + \dots + d_k) + (j_1 + \dots + j_k)
+ (\rho(d_1, d_2) + \dots + \rho(d_1, d_k))
}
[A]_{d_1,j_1}
{\det {\left(
A\binom{d_2, \dots, d_k}{j_2, \dots, j_k}
\right)}}
\det
{(A({d_1,d_2,\dots,d_k}|{j_1,j_2,\dots,j_k}))}
\\
= {} &
\sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n}
\sum_{1 \leq s \leq k}
(-1)^{
(i_1 + \dots + i_k) + (j_1 + \dots + j_k)
+ (s-1)
}
[A]_{i_s,j_1}
{\det {\left(
A\binom{i_1,\dots,i_{s-1},i_{s+1},\dots,i_k}{j_2, \dots, j_k}
\right)}}
\det
{(A({i_1,i_2,\dots,i_k}|{j_1,j_2,\dots,j_k}))}
\\
= {} &
\sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n}
\left(
(-1)^{s-1}
\sum_{1 \leq s \leq k}
[A]_{i_s,j_1}
{\det {\left(
A\binom{i_1,\dots,i_{s-1},i_{s+1},\dots,i_k}{j_2, \dots, j_k}
\right)}}
\right)
(-1)^{
(i_1 + \dots + i_k) + (j_1 + \dots + j_k)
}
\det
{(A({i_1,i_2,\dots,i_k}|{j_1,j_2,\dots,j_k}))}
\\
= {} &
\sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n}
\left(
(-1)^{s+1}
\sum_{1 \leq s \leq k}
[A]_{i_s,j_1}
{\det {\left(
A\binom{i_1,\dots,i_{s-1},i_{s+1},\dots,i_k}{j_2, \dots, j_k}
\right)}}
\right)
(-1)^{
(i_1 + \dots + i_k) + (j_1 + \dots + j_k)
}
\det
{(A({i_1,i_2,\dots,i_k}|{j_1,j_2,\dots,j_k}))}
\\
= {} &
\sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n}
{
\det {\left(
A\binom{i_1, i_2, \dots, i_k}
{j_1, j_2, \dots, j_k}
\right)}
}
(-1)^{
(i_1 + \dots + i_k) + (j_1 + \dots + j_k)
}
\det
{(A({i_1,i_2,\dots,i_k}|{j_1,j_2,\dots,j_k}))}.
\end{aligned}
$$
We have used the hypothesis in the second equality.
Hence by mathematical induction, $P(k)$ is true for $k = 1$, $2$, $3$, $\dots$.
Note that the proof uses nothing more than the usual Laplace expansion.
Best Answer
Not sure if it has a name in the literature, but this is just $\|M^\ast M\|_F^{1/2}=\left(\sum_{i=1}^n\sigma_i(M)^4\right)^{1/4}$, the $4$-norm of the singular values of $M$. In contrast, the Frobenius norm $\|M\|_F=\sqrt{\operatorname{tr}(M^\ast M)}=\left(\sum_{i=1}^n\sigma_i(M)^2\right)^{1/2}$ is the $2$-norm of the singular values.