The claim is true as stated, and can be arrived at using Weyl's criterion which was pointed out in the comments. As I post this answer, there is no consensus on the rate of convergence of the $N$.

We proceed by induction on $k$, the number of polynomials. For $k=1$, either $p_1$ has only rational coefficients or it has at least one irrational. If it has an irrational one, Weyl's criterion shows that $p_1(n)$ is an equidistributed sequence, and thus has infinitely many $N$ for which $||p_1(N)||<\epsilon$. If $p_1$ has only rational coefficients, we can simply choose $N$ to be the product of every coefficient's denominator to get $||p_1(N)||=0$.

Now suppose the theorem holds for some fixed $k$. Consider a polynomial list $p_0, p_1, ..., p_k$ of length $k+1$. If $\{p_0(x),p_1(x),...,p_k(x), 1, x, x^2,...\}$ is $\mathbb{Q}$-linearly independent, then Weyl's criterion on the $k+1$ torus applies and equidistribution gives the existence of infinitely many $N$ for which $||p_i(N)||<\epsilon$ for all $i=0,...,k$.

If, on the other hand, the aforementioned collection is instead $\mathbb{Q}$-linearly dependent, Weyl's criterion does not apply. However, $$p_0(x)=q_1p_1(x)+\cdots +q_kp_k(x)+r_0+r_1x+\cdots +r_sx^s$$ for $q_i,r_j\in \mathbb{Q}$. If we let $q$ be the product of all denominators of the $r_j$, we have that at any natural number $n$, $$||p_0(qn)||=||p_1(qn)+\cdots +p_k(qn)||.$$

Apply the inductive hypothesis to $p_1(qx),...,p_k(qx)$. Choose $M$ so that $||p_i(qM)||<\epsilon/2k$ for all $i=1,...,k$. At this small scale, the $||\cdot||$ function is subadditive in $k$ arguments, so $$||p_0(qM)||\leq ||p_1(qM)||+\cdots +||p_k(qM)||<k(\epsilon/2k)<\epsilon.$$

Therefore, at $N=qM$, $||p_i(N)||<\epsilon$ for all $i=0,...,k$. We are finished by induction.

The claim is true as stated, and can be arrived at using Weyl's criterion which was pointed out in the comments. As I post this answer, there is no consensus on the rate of convergence of the $N$.

We proceed by induction on $k$, the number of polynomials. For $k=1$, either $p_1$ has only rational coefficients or it has at least one irrational. If it has an irrational one, Weyl's criterion shows that $p_1(n)$ is an equidistributed sequence, and thus has infinitely many $N$ for which $||p_1(N)||<\epsilon$. If $p_1$ has only rational coefficients, we can simply choose $N$ to be the product of every coefficient's denominator to get $||p_1(N)||=0$.

Now suppose the theorem holds for some fixed $k$. Consider a polynomial list $p_0, p_1, ..., p_k$ of length $k+1$. If $\{p_0(x),p_1(x),...,p_k(x), 1, x, x^2,...\}$ is $\mathbb{Q}$-linearly independent, then Weyl's criterion on the $k+1$ torus applies and equidistribution gives the existence of infinitely many $N$ for which $||p_i(N)||<\epsilon$ for all $i=0,...,k$.

If, on the other hand, the aforementioned collection is instead $\mathbb{Q}$-linearly dependent, Weyl's criterion does not apply. However, $$p_0(x)=q_1p_1(x)+\cdots +q_kp_k(x)+r_0+r_1x+\cdots +r_sx^s$$ for $q_i,r_j\in \mathbb{Q}$. If we let $q$ be the product of all denominators of the $r_j$, we have that at any natural number $n$, $$||p_0(qn)||=||q_1p_1(qn)+\cdots +q_kp_k(qn)||.$$

Apply the inductive hypothesis to $q_1p_1(qx),...,q_kp_k(qx)$. Choose $M$ so that $||q_ip_i(qM)||<\epsilon/2k$ for all $i=1,...,k$. At this small scale, the $||\cdot||$ function is subadditive in $k$ arguments, so $$||p_0(qM)||\leq ||q_1p_1(qM)||+\cdots +||q_kp_k(qM)||<k(\epsilon/2k)<\epsilon.$$

Therefore, at $N=qM$, $||p_i(N)||<\epsilon$ for all $i=0,...,k$. We are finished by induction.