There are $10^{12}$ ways to assign the balls to the cells, each of which we assume is equally likely. (It's easy to get off on the wrong foot in this problem by overlooking this assumption, for example by assuming that all non-negative integer solutions to $x_1+x_2+ \dots +x_{10} = 12$ are equally likely, which is a very unlikely assumption.) We want to count all the assignments in which exactly one cell is empty.
To do this, we will apply a variation of the Principle of Inclusion/Exclusion (PIE). Let's say an assignment of balls to cells has "property $i$" if cell $i$ is empty, for $i = 1,2,3,\dots ,10$, so our goal is to find the number of arrangements with exactly one of the properties. Further, we define $S_j$ as the total number of assignments which have $j$ of the properties (with over-counting), for $j = 1, 2, 3, \dots ,9$. Then we have
$$S_j = \binom{10}{j} (10-j)^{12}$$
for $1 \le j \le 9$, since there are $\binom{10}{j}$ ways to pick the empty cells and $(10-j)^{12}$ ways to assign the balls to the remaining cells.
The variation of PIE we want to apply is that if there are $n$ properties, then the number of arrangements with exactly $m$ of the properties is
$$N_m = S_m - \binom{m+1}{m} S_{m+1} + \binom{m+2}{m} S_{m+2} - \dots + (-1)^{n-m} \binom{n}{m}S_n$$
(Reference: Applied Combinatorics, Second Edition by Allan Tuckser, Section 8.2, Theorem 2; or An Introduction to Probability Theory and Its Applications, Volume I, Third Edition, by William Feller, Section IV.3.) The case we are interested in is $m=1$, $n=9$, so
$$N_1 = S_1 - \binom{2}{1} S_2 + \binom{3}{1} S_3 - \dots + \binom{9}{1}S_9$$
which yields $N_1 \approx 8.08315 \times 10^{10}$. Therefore the probability of having exactly one cell empty is
$$\frac{N_1}{10^{12}} \approx \boxed{0.0808315}$$
