jmc

Because the polynomial has rational coefficients, the radical conjugate of each of the given roots must also be roots of the polynomial. However, $1+\sqrt{7}$ and $1-\sqrt{7}$ are each other's radical conjugates, so we only get $2$ more roots. (You might be tempted to think that $3-2\sqrt{2}$ and $-3-2\sqrt{2}$ are also a pair of radical conjugates, but the radical conjugate of $3-2\sqrt{2}$ is $3+2\sqrt{2},$ while the radical conjugate of $-3-2\sqrt{2}$ is $-3+2\sqrt{2}.$ Therefore, each one of the numbers $3-2\sqrt{2}$ and $-3-2\sqrt{2}$ is actually the negation of the radical conjugate of the other one.) In total, the polynomial must have at least $4+2=6$ roots.

Furthermore, the polynomial $$(x-3+2\sqrt{2})(x-3-2\sqrt{2})(x+3+2\sqrt{2})(x+3-2\sqrt{2})(x-1-\sqrt{7})(x-1+\sqrt{7})=(x^2-6x+1)(x^2+6x+1)(x^2-2x-6)$$has roots $3\pm 2\sqrt{2},$ $-3\pm 2\sqrt{2},$ and $1\pm \sqrt{7},$ and has rational coefficients. Hence, the smallest possible degree is $\boxed{6}.$