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
$$\frac{N_1}{10^{12}} \approx \boxed{0.0808315}$$
The problem is that you haven't considered the order of the balls. It's true that there are $5$ patterns (i.e. just counting how many balls are in each cell) with a triple and $10$ patterns without a triple, but each of the patterns with a triple occurs in $\binom 73\times 4!$ ways, since there are this many options for which three balls go in the triple and order the rest. Similarly each pattern without a triple occurs in $\binom 72\times \binom 52\times 3!$ ways.
If we distribute each ball at random, it is the ways with balls labelled which are equally likely, not the patterns.
Best Answer
There are $n^n$ equally likely ways to distribute the $n$ balls among the cells. We find the number of ways in which precisely one cell is empty.
The empty cell can be chosen in $n$ ways. For each such choice, the cell that will have two balls can be chosen in $n-1$ ways.
For each such choice, the two balls that go into the lucky cell can be chosen in $\binom{n}{2}$ ways. That leaves $n-2$ balls and $n-2$ cells. The balls can be permuted among these cells in $(n-2)!$ ways.
So with a denominator of $n^n$, the numerator is $$\binom{n}{1}\binom{n-1}{1}\binom{n}{2}(n-2)!.$$
This simplifies to $\binom{n}{2}n!$.
Remark: For fun, we obtain $\binom{n}{2}n!$ "directly." The balls that are to be glued together can be chosen in $\binom{n}{2}$ ways. Now we have $n$ "abstract" balls: the $n-2$ remaining ordinary balls, our glued pair, and the invisible ball. These $n$ objects are to be assigned, one object to a cell, to our $n$ cells. That can be done in $n!$ ways.