I've found a proper approximation after some time and some searching.

Since the binomial distribution has a very large n, we can use the central limit theorem and treat it as a normal distribution. The mean would be obviously 500 billion, the standard deviation is √(n * p * (1-p)) which results in 500,000.

You still cannot plug that into WA unfortunately so we have to use a workaround.

You would calculate it manually through:

Φ(b) - Φ(a), with
b = (510 billion - mean) / (standard deviation) = 20,000
and
a = (490 billion - mean) / (standard deviation) = -20,000
and
Φ(x) = 0.5 * (1 + erf(x/√2))

erf(x) is the error function which has the neat property: erf(-x) = -erf(x)

You could replace erf(x) with an integral but this would be illegible without LaTeX.

Therefore:

Φ(20,000) - Φ(-20,000)
= 0.5 * [ erf(20,000/√2) - erf(-20,000/√2) ]
= erf(20,000/√2)
≈ erf(14,142)

WolframAlpha will unfortunately not calculate this either.

However, according to Wikipedia an approximation exists which shows that:

1 - erf(x) ≈ [(1 - e^(-Ax))e^(-x²)] / (Bx√π)

And apparently A = 1.98 and B = 1.135 give good approximations for all x≥0.

After failing to get a proper approximation from WA again and having to calculate every part by itself, the result is very roughly around 1 - 10^(-86,857,234).

So it is very safe to assume you will lose between 49% and 51% of your gut bacteria. For a more realistic 10 trillion you should replace a and b above with around ±63,200 but I don't want to bother calculating the rest and having WolframAlpha tell me my intermediary steps are equal to zero.