If $I$ is the ideal over integers mod 4 generated by $2x^2+2$ then $I$ has no monic polynomials of degree 2.
Note I'm indebted to Daniel Fisher's comment to choose mod 4 rather than mod 3 as I had earlier.
An element of the ideal $I$ is obtained by multiplying $2x^2+2$ either by a polynomial of degree at least one (which makes the degree 3 or more) or else by multiplying $2x^2+2$ by an element of the integers mod 4. Since $2$ has no inverse mod $4$ we cannot choose a constant so as to multiply $2x^2+2$ by it, and obtain a monic polynomial of degree 2. This shows that this $I$ has no monic polynomials of degree 2.
By the way I am taking $n$ as the lowest positive degree. If we allow degree zero, i.e. nonzero constants, as the generating polynomials, we can take the case of the single element $2$ in the polynomials over the integers mod $4$, and generate the ideal $I=\{0,2\}$ as an example, which is an ideal of only constant polynomials, and here $n=0$ and there is no monic polynomial of degree $0$ in $I$ since that monic would have to be $1$.