Brainstorming:
As you pointed out. If one of the digits of $N$ is $7$ then $7! > 1000> N$ which is impossible.
If one of the digits is $6$ then $N > 6! = 720$ so one of the digits is at least $7$ which we just showed was impossible.
Also if all three digits, call them $a,b,c$, are all $< 5$ then $a!, b!,c! \le 4! =24$ and $N = a! + b!+c! \le 3*24 = 72 < 100$ so that's impossible. So there must be at least one digit that is equal to $5$.
So one of the digits is $5$. Then if the other two are $a,b$ then $a,b \le 5$ and $N = a! + b! + 5! \le 5! + 5! + 5! = 360$. So the first digit is $3$ or less.
So $N = a! + b! +5! \le 3! + 5! + 5! = 246$ so the first digit is $2$ or less.
If the first digit, let's assume the first digit is $a$ is equal $2$ then $200 < N = 2! + b! + 5! = b! + 122$ so $b! > 78$ so $b >4$ but $b \le 5$ so $b=5$ and $N = 255$ but $2! + 5! + 5! = 242$ so $255$ is not factorial. So the first digit is less than $2$.
But $N > 100$ so the first digit is $1$.
So $N = 1! + b! + 5! = 126 +b!$. Where $b = 0,1,2,3,4,5$ and I guess at this point we can do trial and error.
But we have $100 < N < 200$ and $N = 126+b!$ so $b! < 74$ for $b \le 4$. So $N=126 +b! \le 126 + 24 = 150$. And $1! + 5! + 0! = 122\ne 150$ so $150$ is not factorial so $N \le 149$. but as one of the digits is $5$ that must be the last digit and $N = 100 + 10b + 5$.
So $N = 105 + 10b = 1! + b! + 5! = 121 + b!$ so $10b = b! + 16$. And we know $N>121$ so $b \ge 2$ and $b \le 4$ and $b! + 16 \equiv 0 \pmod {10}$ and.... how much further can I go without saying, look, $b$ has got to be $4$; that's the only option?
Well a little further.... just to be masochistic. $10b = 20,30$ or $40$ so $b! = 4,14$ or $24$ and $b = 2,3,4$ so either $2! = 4$, $3!=14$ or $4! = 24$ and ....
That's as far as I can go.
$b =4$ and $N = 100 + 40 + 5 = 1! + 4! +5!$.