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As mentioned by Will Sawin, a necessary condition is that $p$ divides $n$. Thus let us assume that $n=pk$. Denoting $e_1,\dotsc,e_n$ the canonical basis, the knowledge of the $p\times p$ minors is the knowledge of the $p$-vectors $$(Ae_{i_1})\wedge\cdots\wedge(Ae_{i_p})\in\Lambda^p(K),$$ where $K$ is the field of scalars (e.g. $\mathbb C$).

Splitting $$(Ae_1)\wedge\cdots\wedge (Ae_n) = [(Ae_1)\wedge\cdots\wedge (Ae_p)]\wedge\cdots\wedge[(Ae_{n-p+1})\wedge\cdots\wedge (Ae_n)], $$ we see that $(Ae_1)\wedge\cdots\wedge (Ae_n)$ is a polynomial function in the $p\times p$ minors. Since $(Ae_1)\wedge\dotsb\wedge (Ae_n)=(\det A)e_1\wedge\dotsb\wedge e_n$, we deduce the value of $\det A$.

Let me describe how it works when $n=4$ and $p=2$. The minors are denoted $$A\binom{i\alpha}{j\beta}=a_{i\alpha}a_{j\beta}-a_{i\beta}a_{j\alpha}.$$ Then $$(Ae_\alpha)\wedge(Ae_\beta)=\sum_{i<j}A\binom{i\alpha}{j\beta}e_i\wedge e_j.$$ We thus obtain $$\det A=\sum_{\substack{\rho\in{\frak S}_4 \\ \rho(1)<\rho(2),\rho(3)<\rho(4)}}A\binom{\rho(1)1}{\rho(2)2}A\binom{\rho(3)3}{\rho(4)4}.$$

As mentioned by Will Sawin, a necessary condition is that $p$ divides $n$. Thus let us assume that $n=pk$. Denoting $e_1,\dotsc,e_n$ the canonical basis, the knowledge of the $p\times p$ minors is the knowledge of the $p$-vectors $$(Ae_{i_1})\wedge\cdots\wedge(Ae_{i_p})\in\Lambda^p(K^n),$$ where $K$ is the field of scalars (e.g. $\mathbb C$).

Splitting $$(Ae_1)\wedge\cdots\wedge (Ae_n) = [(Ae_1)\wedge\cdots\wedge (Ae_p)]\wedge\cdots\wedge[(Ae_{n-p+1})\wedge\cdots\wedge (Ae_n)], $$ we see that $(Ae_1)\wedge\cdots\wedge (Ae_n)$ is a polynomial function in the $p\times p$ minors. Since $(Ae_1)\wedge\dotsb\wedge (Ae_n)=(\det A)e_1\wedge\dotsb\wedge e_n$, we deduce the value of $\det A$.

Let me describe how it works when $n=4$ and $p=2$. The minors are denoted $$A\binom{i\alpha}{j\beta}=a_{i\alpha}a_{j\beta}-a_{i\beta}a_{j\alpha}.$$ Then $$(Ae_\alpha)\wedge(Ae_\beta)=\sum_{i<j}A\binom{i\alpha}{j\beta}e_i\wedge e_j.$$ We thus obtain $$\det A=\sum_{\substack{\rho\in{\frak S}_4 \\ \rho(1)<\rho(2),\rho(3)<\rho(4)}}A\binom{\rho(1)1}{\rho(2)2}A\binom{\rho(3)3}{\rho(4)4}.$$

As mentioned by Will Sawin, a necessary condition is that $p$ divides $n$. Thus let us assume that $n=pk$. Denoting $e_1,\dotsc,e_n$ the canonical basis, the knowledge of the $p\times p$ minors is the knowledge of the $p$-vectors $$(Ae_{i_1})\wedge\cdots\wedge(Ae_{i_p})\in\Lambda^p(K),$$ where $K$ is the field of scalars (e.g. $\mathbb C$).

Splitting $$(Ae_1)\wedge\cdots\wedge (Ae_n) = [(Ae_1)\wedge\cdots\wedge (Ae_p)]\wedge\cdots\wedge[(Ae_{n-p+1})\wedge\cdots\wedge (Ae_n)], $$ we see that $(Ae_1)\wedge\cdots\wedge (Ae_n)$ is a polynomial function in the $p\times p$ minors. Since $(Ae_1)\wedge\dotsb\wedge (Ae_n)=(\det A)e_1\wedge\dotsb\wedge e_n$, we deduce the value of $\det A$.

Let me describe how it works when $n=4$ and $p=2$. The minors are denoted $$A\binom{i\alpha}{j\beta}=a_{i\alpha}a_{j\beta}-a_{i\beta}a_{j\alpha}.$$ Then $$(Ae_\alpha)\wedge(Ae_\beta)=\sum_{i<j}A\binom{i\alpha}{j\beta}e_i\wedge e_j.$$ We thus obtain $$\det A=\sum_{\rho\in{\frak S}_4;\rho(1)<\rho(2),\rho(3)<\rho(4)}A\binom{\rho(1)1}{\rho(2)2}A\binom{\rho(3)3}{\rho(4)4}.$$

As mentioned by Will Sawin, a necessary condition is that $p$ divides $n$. Thus let us assume that $n=pk$. Denoting $e_1,\dotsc,e_n$ the canonical basis, the knowledge of the $p\times p$ minors is the knowledge of the $p$-vectors $$(Ae_{i_1})\wedge\cdots\wedge(Ae_{i_p})\in\Lambda^p(K),$$ where $K$ is the field of scalars (e.g. $\mathbb C$).

Splitting $$(Ae_1)\wedge\cdots\wedge (Ae_n) = [(Ae_1)\wedge\cdots\wedge (Ae_p)]\wedge\cdots\wedge[(Ae_{n-p+1})\wedge\cdots\wedge (Ae_n)], $$ we see that $(Ae_1)\wedge\cdots\wedge (Ae_n)$ is a polynomial function in the $p\times p$ minors. Since $(Ae_1)\wedge\dotsb\wedge (Ae_n)=(\det A)e_1\wedge\dotsb\wedge e_n$, we deduce the value of $\det A$.

Let me describe how it works when $n=4$ and $p=2$. The minors are denoted $$A\binom{i\alpha}{j\beta}=a_{i\alpha}a_{j\beta}-a_{i\beta}a_{j\alpha}.$$ Then $$(Ae_\alpha)\wedge(Ae_\beta)=\sum_{i<j}A\binom{i\alpha}{j\beta}e_i\wedge e_j.$$ We thus obtain $$\det A=\sum_{\substack{\rho\in{\frak S}_4 \\ \rho(1)<\rho(2),\rho(3)<\rho(4)}}A\binom{\rho(1)1}{\rho(2)2}A\binom{\rho(3)3}{\rho(4)4}.$$

As mentioned by Will Sawin, a necessary condition is that $p$ divides $n$. Thus let us assume that $n=pk$. Denoting $e_1,\ldots,e_n$ the canonical basis, the knowledge of the $p\times p$ minors is the knowledge of the $p$-vectors $$(Ae_{i_1})\wedge\cdots\wedge(Ae_{i_p})\in\Lambda^p(K),$$ where $K$ is the field of scalars (e.g. $\mathbb C$).

Splitting $$(Ae_1)\wedge\cdots\wedge (Ae_n) = [(Ae_1)\wedge\cdots\wedge (Ae_p)]\wedge\cdots\wedge[(Ae_{n-p+1})\wedge\cdots\wedge (Ae_n)], $$ we see that $(Ae_1)\wedge\cdots\wedge (Ae_n)$ is a polynomial function in the $p\times p$ minors. Since $(Ae_1)\wedge\cdots\wedge (Ae_n)=(\det A)e_1\wedge\cdots\wedge e_n$, we deduce the value of $\det A$.

Let me describe how it works when $n=4$ and $p=2$. The minors are denoted $$A\binom{i\alpha}{j\beta}=a_{i\alpha}a_{j\beta}-a_{i\beta}a_{j\alpha}.$$ Then $$(Ae_\alpha)\wedge(Ae_\beta)=\sum_{i<j}A\binom{i\alpha}{j\beta}e_i\wedge e_j.$$ We thus obtain $$\det A=\sum_{\rho\in{\frak S}_4;\rho(1)<\rho(2),\rho(3)<\rho(4)}A\binom{\rho(1)1}{\rho(2)2}A\binom{\rho(3)3}{\rho(4)4}.$$

As mentioned by Will Sawin, a necessary condition is that $p$ divides $n$. Thus let us assume that $n=pk$. Denoting $e_1,\dotsc,e_n$ the canonical basis, the knowledge of the $p\times p$ minors is the knowledge of the $p$-vectors $$(Ae_{i_1})\wedge\cdots\wedge(Ae_{i_p})\in\Lambda^p(K),$$ where $K$ is the field of scalars (e.g. $\mathbb C$).

Splitting $$(Ae_1)\wedge\cdots\wedge (Ae_n) = [(Ae_1)\wedge\cdots\wedge (Ae_p)]\wedge\cdots\wedge[(Ae_{n-p+1})\wedge\cdots\wedge (Ae_n)], $$ we see that $(Ae_1)\wedge\cdots\wedge (Ae_n)$ is a polynomial function in the $p\times p$ minors. Since $(Ae_1)\wedge\dotsb\wedge (Ae_n)=(\det A)e_1\wedge\dotsb\wedge e_n$, we deduce the value of $\det A$.

Let me describe how it works when $n=4$ and $p=2$. The minors are denoted $$A\binom{i\alpha}{j\beta}=a_{i\alpha}a_{j\beta}-a_{i\beta}a_{j\alpha}.$$ Then $$(Ae_\alpha)\wedge(Ae_\beta)=\sum_{i<j}A\binom{i\alpha}{j\beta}e_i\wedge e_j.$$ We thus obtain $$\det A=\sum_{\rho\in{\frak S}_4;\rho(1)<\rho(2),\rho(3)<\rho(4)}A\binom{\rho(1)1}{\rho(2)2}A\binom{\rho(3)3}{\rho(4)4}.$$