Python 3, 25 bytes

lambda n:len(bin(n**n))-3

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If the answer is \$x\$, then \$x+1 > n\log{n} >= x\$ holds true, which means \$2^{x+1} > n^n >= 2^x\$. So we can simply count the number of bits in the binary representation of \$n^n\$.