JavaScript (ES6), 153 142 139 bytes
n=>([...n].map((c,i,[...a])=>[...''+1e9].map((u,j)=>s+=j+i&&j!=c?p((a.splice(i,1,j),a.join``)):0),s=0,p=q=>eval('for(k=q;q%--k;);k==1')),s)
Accepts input as a string. Undefined behavior for invalid input, though it should terminate without error on any string I can think of. Not necessarily before the heat-death of the universe though, particularly for long strings.
Demo
f=
n=>([...n].map((c,i,[...a])=>[...''+1e9].map((u,j)=>s+=j+i&&j!=c?p((a.splice(i,1,j),a.join``)):0),s=0,p=q=>eval('for(k=q;q%--k;);k==1')),s)
console.log([...''+1e19].map((_,i)=>f(i+1+'')).join())
i.onchange=()=>console.log(f(i.value))
<input id=i>
Improvements
Saved 11 bytes by refactoring the reduce() calls into map() calls, and by implicitly copying the array a in the function parameter, instead of in within the context of the splice() call.
Saved 3 bytes thanks to @Neil's suggestion to convert [...Array(10)] to [...''+1e9].
Unminified code
input => (
[...input].map(
(char, decimal, [...charArray]) =>
[...'' + 1e9].map(
(unused, digit) => sum +=
digit + decimal && digit != char ?
prime(
(
charArray.splice(decimal, 1, digit)
, charArray.join``
)
) :
0
)
, sum = 0
, prime = test => eval('for(factor = test; test % --factor;); factor == 1')
)
, sum
)
Explanation
The function uses a two-level map() to sum the amount of permutations that pass the primality test, which was borrowed and modified from this answer.
(Original answer)
reduce((accumulator, currentValue, currentIndex, array) => aggregate, initialValue)
So for example, to calculate the sum of an array, you would pass an initialValue of 0, and return an aggregate equal to accumulator + currentValue. Modifying this approach slightly, we instead calculate the number of permutations that pass the primality test:
reduce(
(passedSoFar, currentDecimal, currentIndex, digitArray) =>
isValidPermutation() ?
passedSoFar + prime(getPermutation()) :
passedSoFar
, 0
)
That is essentially the inner reduce(), which iterates all the permutations of the digitArray by changing each decimal to a specific permutatedDigit. We then need an outer reduce() to iterate all possible permutatedDigit's with which to replace each decimal, which is just 0-9.
Abnormalities in implementation
[...''+1e9].map((u,j)=>... was the shortest way @Neil could think of to iterate an argument 0 through 9. It would be preferable to do so with u, but u is not useful for each element in the array, in this case.
i+j in the ternary condition checks to ensure that 0 is not a possible permutation of the leading digit, as per the challenge specification. j!=c ensures that the original n is not a candidate to go through the primality test.
(a.splice(i,1,j),a.join``) is kind of a mess. splice() replaces the digit at decimal == i with the permutatedDigit == j, but since splice() returns the removed elements (in this case, would be equal to [a[i]]) instead of the modified array, we must use the comma operator to pass the modified array a to the primality test, but not before join()ing it into a number string.
Lastly, the eval() is to save a byte since, compared to the more canonical approach, it is shorter:
q=>eval('for(k=q;q%--k;);k==1')
q=>{for(k=q;q%--k;);return k==1}
The reference to the prime test p is initialized in an unused argument to the map() call.
nfor which the output is0. I think it'sn = 200. I also think they come in bunches:200,202,204,206,208,320,322,...,328,510,...,518,620,...628,840,...,848, etc. \$\endgroup\$