What is the probability on rolling $2n$ dice that the sum of the first $n$ equals the sum of the last $n$?

I don't know whether you're interested in approximate and asymptotic answers – there's a straightforward estimate for large $n$. The distribution for the sum tends to a normal distribution. The variance for one die is

$$ \langle x^2\rangle-\langle x\rangle^2=\frac{1+4+9+16+25+36}6-\left(\frac{1+2+3+4+5+6}6\right)^2=\frac{35}{12}\;, $$

so the variance for $n$ dice is $n$ times that. The probability of a tie is the sum over the squares of the probabilities, which we can approximate by the integral over the square of the density, so this is

$$ \int_{-\infty}^\infty\left(\frac{\exp\left(-x^2/\left(2\cdot\frac{35}{12}n\right)\right)}{\sqrt{2\pi\frac{35}{12}n}}\right)^2\,\mathrm dx=\int_{-\infty}^\infty\frac{\exp\left(-x^2/\left(\frac{35}{12}n\right)\right)}{2\pi\frac{35}{12}n}\,\mathrm dx=\sqrt{\frac3{35\pi n}}\approx\frac{0.1652}{\sqrt n}\;. $$

The approximation is already quite good for $n=3$, where it yields about $0.095$ whereas your exact answer is about $0.093$.

It is possible to calculate this exactly if you are willing to use arbitrary precision integer arithmetic.

You can use the recursion $$f(n,k)=\sum_{j=1}^6 f(n-1,k-j)$$ starting at $f(0,0)=1$ and $f(0,k)=0$ when $k\not =0$ to find the number of ways of scoring $k$ from $n$ dice. Your result is then $$\sum_{i=n}^{6n} f(n,i)^2 / 6^{2n}$$ which is the division of two very large integers: for $n=200$ the numerator will be about $2.1\times 10^{309}$ and the denominator will be $6^{400}\approx 1.8\times 10^{311}$.

More practically using a spreadsheet and only looking for several decimal places you can use $$g(n,k)=\sum_{j=1}^6 g(n-1,k-j) / 6$$ starting at $g(0,0)=1$ and $f(0,k)=0$ when $k\not =0$ to find the probability of scoring $k$ from $n$ dice. Your result is then $$\sum_{i=n}^{6n} g(n,i)^2.$$

With $n=200$ this latter method will just over 200 columns and 1200 rows of the spreadsheet, so not difficult, and an extra column for the squares of the final column. In practice it give a value of about $0.0116752$ for the probability of matched sums rolling 200 dice twice.

This compares with about $0.0116798$ from joriki's approximation, a relative difference of around 0.04%.

This is not as explicit as you may want, but it's a first step. I encourage you to read about generating function, one possible answer to your question lies in them. (Plus they are really fun!).

One can show that the number of ways to write $k=x_1+x_2+\cdots +x_n$ where $1\leq x_i\leq 6$ for all $i$ are the coefficient of $x^k$ in $$ (x+x^2+x^3+x^4+x^5+x^6)^n. $$ This coefficient is the multinomial coefficient $${n\choose x_1,x_2,\cdots, x_m}$$, where $k=x_1+x_2+\cdots +x_n$ for a given $k$.

Now what you want to do is sum the square of these coefficients and divide it by $36^n$

We can write this as $$ \frac{1}{36^n}\sum_{k=n}^{6n}\quad\sum_{x_1+x_2+\cdots +x_n=k}{n\choose x_1,x_2,\cdots, x_m}^2 $$ Substituting $n=2$ we get $\frac{146}{1296}$ and for $n=3$ we get $ \frac{4332}{46656}.$