If $e_1,\ldots,e_N$ are the elementary symmetric functions of $x_1,\ldots,x_N$ then for $k\ge N$ one has $$p_k=\sum_{j=1}^N(-1)^{j-1}e_j p_{k-j}.$$ This formula uses the elementary symmetric functions, which I presume you want to avoid, but it means that for $k\ge 2N$ the $(N+1)$ by $(N+1)$ matrix $$M_k=(p_{k-i-j})_{i,j=0}^N$$ has the null-vector $(1,-e_1,e_2,-e_3,\ldots,\pm e_N)$ and so $\det(M_k)=0$. Expanding this out gives an explicit formula for $p_k$ as a rational function (alas!) of $p_{k-1},\ldots,p_{k-2N}$.

If $e_1,\ldots,e_N$ are the elementary symmetric functions of $x_1,\ldots,x_N$ then for $k\ge N$ one has $$p_k=\sum_{j=1}^N(-1)^{j-1}e_j p_{k-j}.$$ This formula uses the elementary symmetric functions, which I presume you want to avoid, but it means that for $k\ge 2N$ the $(N+1)$ by $(N+1)$ matrix $$M_k=(p_{k-i-j})_{i,j=0}^N$$ has the null-vector $(1,-e_1,e_2,-e_3,\ldots,\pm e_N)$ and so $\det(M_k)=0$. Expanding this out gives an explicit formula for $p_k$ as a rational function (alas!) of $p_{k-1},\ldots,p_{k-2N}$.

Added I suppose one can express the $e_j$ in terms of $p_1,\ldots,p_n$ and put them into the above linear recurrence for $p_k$.

If $e_1,\ldots,e_N$ are the elementary symmetric functions of $x_1,\ldots,x_N$ then for $k\ge N$ one has $$p_k=\sum_{j=1}^N(-1)^{j-1}e_j p_{k-j}.$$ This formula uses the elementary symmetric functions, which I presume you want to avoid, but it means that $k\ge 2N$ the matrix $(N+1)$ by $(N+1)$ matrix $$M_k=(p_{k-i-j})_{i,j=0}^N$$ has the null-vector $(1,-e_1,e_2,-e_3,\ldots,\pm e_N)$ and so $\det(M_k)=0$. Expanding this out gives an explicit formula for $p_k$ as a rational function (alas!) of $p_{k-1},\ldots,p_{k-2N}$.

If $e_1,\ldots,e_N$ are the elementary symmetric functions of $x_1,\ldots,x_N$ then for $k\ge N$ one has $$p_k=\sum_{j=1}^N(-1)^{j-1}e_j p_{k-j}.$$ This formula uses the elementary symmetric functions, which I presume you want to avoid, but it means that for $k\ge 2N$ the $(N+1)$ by $(N+1)$ matrix $$M_k=(p_{k-i-j})_{i,j=0}^N$$ has the null-vector $(1,-e_1,e_2,-e_3,\ldots,\pm e_N)$ and so $\det(M_k)=0$. Expanding this out gives an explicit formula for $p_k$ as a rational function (alas!) of $p_{k-1},\ldots,p_{k-2N}$.