Simplifying boolean algebra without using certain operations

You could try giving Simplify a ComplexityFunction option that discourages Xor:

Simplify[
    !(x&&y&&z)&&!(x&&!y&&!z)&&!(!x&&!y&&z),
    ComplexityFunction->(LeafCount[#]+10000Count[#, _Xor,{0,Infinity}]&)
]

(x && ! y && z) || (! x && (y || ! z)) || (y && ! z)

Addendum

(The OP asked about another expression)

I don't know which boolean transformations are built in to Simplify/FullSimplify, but you can add more with the option TransformationFunctions. So:

FullSimplify[
    !u && !v && !w && !x,
    ComplexityFunction -> (LeafCount[#]+10000Count[#, _Xor,{0,Infinity}]&),
    TransformationFunctions -> {Automatic, BooleanConvert[#,"OR"]&}
]

(* !(u || v || w || x) *)

perhaps BooleanMinimize?

BooleanMinimize[! (x && y && z) && ! (x && ! y && ! z) && ! (! x && ! y && z)]

(x && ! y && z) || (! x && y) || (! x && ! z) || (y && ! z)

BooleanMinimize[! (x && y && z) && ! (x && ! y && ! z) && ! (! x && ! y && z), "CNF"]

(! x || ! y || ! z) && (! x || y || z) && (x || y || ! z)