I am fairly sure this is a studied problem, but I am no expert. We have a ballot with $k$ candidates and $n \geq k$ paritcipants, each participant voting once, each candidate having equal strength, so that the individual votes are uniformly distributed.
I would like to calculate the probability of the ballot ending in a tie, and possibly to study how for fixed $k$ this changes with $n$ (and asympotically with $n$ and $k$, perhaps).
That's how I would approach the problem. The number of possible outcomes is the total number of weak $k$-compositions of $n$, that is ${n+k-1}\choose{k-1} $. Letting $x=(x_1, x_2, \ldots, x_k ) \in \mathbb N^k$ the set of the favourable outcomes $F$ is given by
$$F=\left\{x \in \mathbb N^k \,\Big| \, \sum_{j=1}^kx_j=n, \, \|x\|_{\infty}=x_i, \, i \in K, K \subseteq \{1, \ldots ,k\}, \, \mbox{card(K)} \geq 2 \right \}$$
How do I find the cardinality of $F$?
Let next suppose that each voter has now fixed $l \leq k$ distinct choices instead of a single vote (if repeated choices are allowed we fall back in the problem above). How does the problem change?
Your mistake
First of all, your formulation of the problem is incorrect. To see this, it suffices to look at a small case, with just $k=2$ candidates and $n=2$ participants.
According to you, the number of total outcomes is $\binom{2+2-1}{2-1}=3$. These three outcomes are the three possible vote distributions: either $(2,0)$, $(1,1)$, or $(0,2)$. Only the distribution of $(1,1)$ results in a tie. There is one favorable outcome out of three total outcomes, so the probability should be $1/3$, right?
Wrong. Since both voters are equally likely to vote for each candidate, and the votes are independent, the probability of the outcome $(2,0)$ where both voters vote for the first candidate is $(1/2)\times (1/2)=1/4$. Similarly, the probability of $(0,2)$ is $1/4$. Therefore, the probability of a tie is $1-1/4-1/4=1/2$.
The problem with your method, based on counting the number of favorable vote distribution, is that not every vote distribution is equally likely, as the last calculation shows.
Approximate result
Let $X_1,\dots,X_k$ be random variables representing the number of votes each candidate receives. The distribution of the random vector $(X_1,\dots,X_k)$ is quite complicated, so I strongly doubt that you will get an exact answer to your question. However, we can approximate this distribution using a technique called Poissonization.
Let $Z_1,\dots,Z_k$ be independent Poisson random variables, each with expected value $n/k$. The random vector $(Z_1,\dots,Z_k)$ is a good approximation for the vector $(X_1,\dots,X_k)$, in the following sense.
Since the expected value of $Z_1+\dots+Z_k$ is $n$, and since $Z_1+\dots+Z_k$ is tightly concentrated around its mean, the effect of conditioning is not too severe, so we conclude that $(Z_1,\dots,Z_k)$ is a good approximation for $(X_1,\dots,X_k)$. This theorem is helpful because while the variables $X_1,\dots,X_k$ are dependent, the variables $Z_1,\dots,Z_k$ are independent, so they are easier to handle.
For the Poissonized version of the event, the probability that there is a tie for largest among the Poisson variables $(Z_1,\dots,Z_k)$ is $$ \begin{aligned} P(\text{tie}) &=1-P(\text{unique winner}) \\&=1-k\sum_{m=0}^\infty P(Z_1=m)\cdot P(Z_1<m)^{k-1} \end{aligned} $$
This is because the probability a particular candidate wins with $m$ points is $P(Z_1=m)\cdot P(Z_1<m)^{k-1}$, because the other $k-1$ candidates need fewer than $m$ points. We then sum over all possible $m$, then multiply by $k$ because any candidate can win. To be clear, $P(Z_1=m)=e^{-n/k}(n/k)^m/m!$, while $P(Z_1<m)=\sum_{r=0}^{m-1}e^{-n/k}(n/k)^r/r!$, so the above formula can be computed easily.
Here are my findings for how well this approximation works, for some sample values of $k$ and $n$.
Here is the code to produce this table: Try it online!