I believe the following values are consistent with the equations:
A = 6, C = 7, L = 3, N = 0, O = 4, S = 5, U = 2, V = 1
My first discovery is that
If L is 4 or greater, one of the remaining digits must be 6 or greater since there are eight of them total, and 6^4 exceeds 1000. Meanwhile, if L is 2 or less, this also does not work because of the second equation, where U^2+V^2+C^2 = 200 + SU. Assuming maximum values for U, V, and C, you would get 81+64+49 = 194, which is insufficient. Therefore, L must be equal to 3.
Working from Equations 1 and 4:
S^3+O^3+3^3 = A^3+N^3, as both equations equal to UVC and the V^L's cancel out. I feel like there may have been a more elegant way to figure this out, but I recall that 3^3 + 4^3 + 5^3 = 6^3, so I believe S and O are 4 and 5 while A and N are 0 and 6.
This gives us:
216 + V^3 = UVC The remaining possibilities for V are 2, 1, 7, 8, and 9, however we can rule out the latter two from equation 2 as 8^3 exceeds 400 and therefore cannot sum to LSU. 7 does not work as you result with 559, therefore V would have to be both 5 and 7. 2 does not work as your result would be 224, and both C and S/O cannot be 4. Lastly V=1 gives you a result of 217, making U = 2, V = 1, and C = 7. So far we are consistent.
Now back to equation 2:
We have 2^3 + 1^3 + 7^3 = LSU, resulting in LSU = 352, giving us L = 3 and U = 2, which is consistent, and S = 5, leaving O to equal 4. We are now left with A and N.
Finally with equation 3:
3^3 + 5^3 + 2^3 = VAN, resulting in VAN = 160. Once again confirming V = 1, as well as A = 6 and N = 0, consistent with our original hypothesis from Equations 1 and 4.