The case of degree $1$ has been handled by quid. This turns out to be an exceptional case, and for all degrees $d\ge 2$ one can show that there is a constant $K(d)$ such that the number of degree $d$ polynomials with coefficient bounded by $C$ and having an integer root is
$$
\sim K(d) (2C+1)^d.
$$
Moreover, for large $d$, the constant $K(d)$ is approximately equal to
$$
1+ 2 \frac{\sqrt{6}}{\sqrt{\pi d}}.
$$
In this approximation, the term $1$ gives the contribution of polynomials having $0$ as a root, and the second term $\sqrt{6/\pi d}$ accounts for polynomials having a root at $1$ (and another similar contribution from those having a root at $-1$). For large $d$, the effect of having roots at integers larger than $1$ in size is substantially smaller.

For $d=10$ and $C=100$ this approximation is about $0.93\%$ which is a little higher than your data, but for $d=9$ it is much closer (the approximation being $0.96\%$). Perhaps the Monte-Carlo simulations haven't fully stabilized?

Clearly the number of polynomials having a root at $0$ is $(2C+1)^d$. Now suppose that $k\ge 1$ is a positive integer, and consider polynomials having a root at $k$ (naturally the same holds for $-k$). Write $f(x)=a_dx^d+\ldots+a_0 = (x-k) (b_{d-1}x^{d-1} + \ldots +b_0)$. Note that $kb_0$ must lie in $[-C,C]$ giving us about $(2C+1)/k$ choices for $b_0$. Next if $b_0$ is fixed, then $-kb_1+b_0$ must lie in $[-C,C]$ giving us about $(2C+1)/k$ choices for $b_1$. Proceeding in this manner, we get at most $(2C+1)/k$ choices for each $b_j$, with the additional constraint that the final $b_{d-1}$ must also be constrained to be in $[-C,C]$. It follows that there are at most $(2C+1)^d/k^d$ possible polynomials having a root at $k$. This upper bound summed over $k$ converges for $d\ge 2$ (but not for $d=1$), and shows that the proportion of polynomials having a large integer root is very small. Thus at any rate the number of polynomials having an integer root is at most $(1+2\zeta(d)) (2C+1)^d$, and by similar reasoning we may see that the number of polynomials having two integer roots is at most $O((2C+1)^{d-1+\epsilon})$.

By our work above, the number of polynomials having an integer root is essentially the sum of those polynomials having a root at $k$ over integers $|k|\le K$ for some slowly growing $K$. Now for a given $k\ge 1$, for the polynomial to have a root at $k$ means that given $a_1$, $\ldots$, $a_d$ (all chosen in $[-C,C]$) we must have the sum $a_1k+a_2k^2+\ldots +a_d k^d$ lying in $[-C,C]$ (which then uniquely determines $a_0$). But now we may write $a_j=Cx_j$, and then the $x_j$ behave like independent random variables chosen uniformly from $[-1,1]$ and then we are asking for the probability that $x_1 k+x_2k^2+\ldots +x_d k^d$ also lies in $[-1,1]$. Clearly this probability must be some constant $K(d,k)$ (which by our earlier work is at most $1/k^d$), and therefore our claimed asymptotic holds with
$$
K(d) = 1+ 2\sum_{k=1}^{\infty} K(d,k).
$$

Lastly we come to the approximation for $K(d)$ for large $d$. Note that the contribution of terms $k\ge 2$ is $O(2^{-d})$ is extremely small. It remains to understand $K(d,1)$ -- the probability that the sum of $d$ independent random variables chosen uniformly from $[-1,1]$ also lies in $[-1,1]$. For large $d$, the sum of $d$ independent random variables is approximately normal with mean $0$, and variance $d/3$. From this our approximation follows. By using Parseval, one can also see that
$$
K(d,1) = \int_{-\infty}^{\infty} \Big(\frac{\sin (\pi x)}{\pi x}\Big)^{d+1} dx,
$$
from which we can calculate, for example, that $K(10,1)=0.4109\ldots$ which is pretty close to the approximation $0.437\ldots$.