The fill volumes per bottle for a certain brand of beer follow a normal distribution with
mean 341 ml and standard deviation 3 ml.

If you buy a twelve-pack of beer, what is the probability that the total volume of beer
will exceed 4.1 liters? (Assume that the 12 bottles can be considered a simple random
sample.)

In this case you should use a normal distribution Yi with parameters :
µ = 341 (mean)
sigma = 3 (standard deviation)

i = 1, 2, 3 ... 12
Yi gives the distribution of the volume of beer in bottle i.

Of course this doesn't reflect exactly the reality (for example Y < 0 has a probability > 0 - even if very very small - whereas no bottle can contain a negative quantity of beer !).
But this model is commonly considered as the best to use in your situation.

If Xi is the standard normal distribution :
p(Yi<a) = p( Xi < (a-µ)/sigma )

Your problem is to find :
p( Y1+Y2+...+Y12 > 4100)

This is the same as :
p( X1 + X2 + ... + X12 > A )
where A = (4100 - 12*341) / 3 = 8/3
and X1, X2, ..., X12 are all independant standard normal distributions.

Their sum Z is also a normal distribution, with parameters :
µ = µ1 + µ2 + ... + µ12 = 0
sigma² = s1² + s2² + ... + s12² = 1² + 1² + ... + 1² = 12
sigma = sqrt(12) = 2*sqrt(3) : standard deviation

p ( Z > A ) = p ( X > (A - µ) / sigma ) = p (X > 8 / 2*sqrt(3) ) = p (X > 4 / sqrt(3) )
and X is a standard normal distribution.

You're now able to find the answer in a table :
~ 1.05%