For Question 1: Once $b-a \geq 4$, the norm cannot get arbitrarily small.

Suppose $p(x)$ has leading term $c_n x^n$. The minimum of $\int_a^b p^2(x) \, dx$ over all polynomials $p(x) = c_n x^n + O(x^{n-1})$ with real coefficients is a multiple of $P_n(l(x))$, where $P_n$ is the $n$-th Legendre polynomial and $l$ is the affine-linear transformation $l(x) = (2x-(a+b))/(b-a)$ that takes $(a,b)$ to $(-1,1)$. Now $P(x/2)$ has leading coefficient $$ \lambda_n := \frac12 \frac34 \frac56 \cdots \frac{2n-1}{2n} \sim \frac1{\sqrt{\pi n}} $$ and norm $\int_{-2}^2 P(x/2)^2 \, dx = 4/(2n+1)$. Thus in general $$ \int_a^b p^2(x) \, dx \geq \frac4{(2n+1)\lambda_n^2} \left(\frac{b-a}{4}\right)^{\!2n+1} c_n^2, $$ with the factor $r_n := 4/(2n+1)\lambda_n^2$ approaching $2\pi$ as $n \rightarrow \infty$. Since $|c_n| \geq 1$, we deduce a lower bound on $\int_a^b p^2(x) \, dx$ that approaches $2\pi$ when $b-a=4$, and increases exponentially with $n$ when $b-a > 4$. [Added later, in connection with the recent MO Question 188807: since $\{ r_n \}_{n \geq 0}$ is an increasing sequence and the $n=0$ bound $b-a$ is attained by the integer polynomial $p(x) = 1$ (or $p(x) = -1$), it also follows that $b-a$ is the minimum over all nonzero $p \in {\bf Z}[X]$ once $b-a \geq 4$.]

It follows also that if $b-a>4$ then for any $M$ there are only finitely many $p \in {\bf Z}[X]$ for which $\int_a^b p^2(x) \, dx < M$. When $b-a = 4$ this is not true when $a,b$ are integers; I do not know what happens otherwise. [For the integer case: it is enough to prove it for $(a,b) = (-2,+2)$. let $T_n$ be the $n$-th Chebyshev polynomial; then $p(x) = 2T_n(x/2)$ has integer coefficients and $$ \int_{-2}^2 p(x)^2 \, dx \leq \int_{-2}^2 2^2 \, dx = 16. $$ (In fact for large $n$ the integral approaches $8$.)]

For Question 1: Once $b-a \geq 4$, the norm cannot get arbitrarily small.

Suppose $p(x)$ has leading term $c_n x^n$. The minimum of $\int_a^b p^2(x) \, dx$ over all polynomials $p(x) = c_n x^n + O(x^{n-1})$ with real coefficients is a multiple of $P_n(l(x))$, where $P_n$ is the $n$-th Legendre polynomial and $l$ is the affine-linear transformation $l(x) = (2x-(a+b))/(b-a)$ that takes $(a,b)$ to $(-1,1)$. Now $P(x/2)$ has leading coefficient $$ \lambda_n := \frac12 \frac34 \frac56 \cdots \frac{2n-1}{2n} \sim \frac1{\sqrt{\pi n}} $$ and norm $\int_{-2}^2 P(x/2)^2 \, dx = 4/(2n+1)$. Thus in general $$ \int_a^b p^2(x) \, dx \geq \frac4{(2n+1)\lambda_n^2} \left(\frac{b-a}{4}\right)^{\!2n+1} c_n^2, $$ with the factor $r_n := 4/(2n+1)\lambda_n^2$ approaching $2\pi$ as $n \rightarrow \infty$. Since $|c_n| \geq 1$, we deduce a lower bound on $\int_a^b p^2(x) \, dx$ that approaches $2\pi$ when $b-a=4$, and increases exponentially with $n$ when $b-a > 4$. [Added later, in connection with the recent MO Question 188807: since $\{ r_n \}_{n \geq 0}$ is an increasing sequence and the $n=0$ bound $b-a$ is attained by the integer polynomial $p(x) = 1$ (or $p(x) = -1$), it also follows that $b-a$ is the minimum over all nonzero $p \in {\bf Z}[X]$ once $b-a \geq 4$.]

It follows also that if $b-a>4$ then for any $M$ there are only finitely many $p \in {\bf Z}[X]$ for which $\int_a^b p^2(x) \, dx < M$. When $b-a = 4$ this is not true when $a,b$ are integers; I do not know what happens otherwise. [For the integer case: it is enough to prove it for $(a,b) = (-2,+2)$. let $T_n$ be the $n$-th Chebyshev polynomial; then $p(x) = 2T_n(x/2)$ has integer coefficients and $$ \int_{-2}^2 p(x)^2 \, dx \leq \int_{-2}^2 2^2 \, dx = 16. $$ (In fact for large $n$ the integral approaches $8$.)]

For Question 1: Once $|b-a| \geq 4$, the norm cannot get arbitrarily small.

Suppose $p(x)$ has leading term $c_n x^n$. The minimum of $\int_a^b p^2(x) \, dx$ over all polynomials $p(x) = c_n x^n + O(x^{n-1})$ with real coefficients is a multiple of $P_n(l(x))$, where $P_n$ is the $n$-th Legendre polynomial and $l$ is the affine-linear transformation $l(x) = (2x-(a+b))/(b-a)$ that takes $(a,b)$ to $(-1,1)$. Now $P(x/2)$ has leading coefficient $$ \lambda_n := \frac12 \frac34 \frac56 \cdots \frac{2n-1}{2n} \sim \frac1{\sqrt{\pi n}} $$ and norm $\int_{-2}^2 P(x/2)^2 \, dx = 4/(2n+1)$. Thus in general $$ \int_a^b p^2(x) \, dx \geq \frac4{(2n+1)\lambda_n^2} \left(\frac{b-a}{4}\right)^{\!2n+1} c_n^2, $$ with the factor $4/(2n+1)\lambda_n^2$ approaching $2\pi$ as $n \rightarrow \infty$. Since $|c_n| \geq 1$, we deduce a lower bound on $\int_a^b p^2(x) \, dx$ that approaches $2\pi$ when $b-a=4$, and increases exponentially with $n$ when $b-a > 4$.

It follows also that if $b-a>4$ then then for any $M$ there are only finitely many $p \in {\bf Z}[X]$ for which $\int_a^b p^2(x) \, dx < M$. When $b-a = 4$ this is not true when $a,b$ are integers; I do not know what happens otherwise. [For the integer case: it is enough to prove it for $(a,b) = (-2,+2)$. let $T_n$ be the $n$-th Chebyshev polynomial; then $p(x) = 2T_n(x/2)$ has integer coefficients and $$ \int_{-2}^2 p(x)^2 \, dx \leq \int_{-2}^2 2^2 \, dx = 16. $$ (In fact for large $n$ the integral approaches $8$.)]

For Question 1: Once $b-a \geq 4$, the norm cannot get arbitrarily small.

Suppose $p(x)$ has leading term $c_n x^n$. The minimum of $\int_a^b p^2(x) \, dx$ over all polynomials $p(x) = c_n x^n + O(x^{n-1})$ with real coefficients is a multiple of $P_n(l(x))$, where $P_n$ is the $n$-th Legendre polynomial and $l$ is the affine-linear transformation $l(x) = (2x-(a+b))/(b-a)$ that takes $(a,b)$ to $(-1,1)$. Now $P(x/2)$ has leading coefficient $$ \lambda_n := \frac12 \frac34 \frac56 \cdots \frac{2n-1}{2n} \sim \frac1{\sqrt{\pi n}} $$ and norm $\int_{-2}^2 P(x/2)^2 \, dx = 4/(2n+1)$. Thus in general $$ \int_a^b p^2(x) \, dx \geq \frac4{(2n+1)\lambda_n^2} \left(\frac{b-a}{4}\right)^{\!2n+1} c_n^2, $$ with the factor $r_n := 4/(2n+1)\lambda_n^2$ approaching $2\pi$ as $n \rightarrow \infty$. Since $|c_n| \geq 1$, we deduce a lower bound on $\int_a^b p^2(x) \, dx$ that approaches $2\pi$ when $b-a=4$, and increases exponentially with $n$ when $b-a > 4$. [Added later, in connection with the recent MO Question 188807: since $\{ r_n \}_{n \geq 0}$ is an increasing sequence and the $n=0$ bound $b-a$ is attained by the integer polynomial $p(x) = 1$ (or $p(x) = -1$), it also follows that $b-a$ is the minimum over all nonzero $p \in {\bf Z}[X]$ once $b-a \geq 4$.]

It follows also that if $b-a>4$ then for any $M$ there are only finitely many $p \in {\bf Z}[X]$ for which $\int_a^b p^2(x) \, dx < M$. When $b-a = 4$ this is not true when $a,b$ are integers; I do not know what happens otherwise. [For the integer case: it is enough to prove it for $(a,b) = (-2,+2)$. let $T_n$ be the $n$-th Chebyshev polynomial; then $p(x) = 2T_n(x/2)$ has integer coefficients and $$ \int_{-2}^2 p(x)^2 \, dx \leq \int_{-2}^2 2^2 \, dx = 16. $$ (In fact for large $n$ the integral approaches $8$.)]